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arV19569 
Elementary calculus 



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ELEMENTARY CALCULUS 



THE MACMILLAN COMPANY 

NEW YORK • BOSTON • CHICAGO • DALLAS 
ATLANTA • SAN FRANCISCO 

MACMILLAN & CO., Limited 

LONDON • BOMBAY ■ CALCUTTA 
MELBOURNE 

THE MACMILLAN CO. OF CANADA, Ltd. 

TORONTO 



ELEMENTARY CALCULUS 



BY 
WILLIAM F. OSGOOD, Ph.D., LL.D 

PEKKIN8 PK0FBS80K OF MATHEMATICS 
IN HABVAKD UNIVERSITY 



Wetogorft 

THE MACMILLAN COMPANY 

1921 



AU HgMe reeemeel 
5 



COPTBIGHT, 1921, 

bt the macmillan company. 



Set up and electrotyped. Published January, 1931. 



J. S. Gushing Co. — Berwick & Smith Co. 
Norwood, Mass., U.S.A. 



PREFACE 

The object of this book is to present the elements of the 
Differential Calculus in a form easily accessible for the under- 
graduate. It is possible, from the very beginning, to illustrate 
the ideas and methods of the Calculus by means of applications 
to physics and geometry, which the student can readily grasp, 
and which will seem to him of interest and value. To do this, 
the stress in the illustrative examples worked in the text must 
be laid first of all on the thought which underlies the method 
of solution, in distinction from the exposition of a process, re- 
duced in the worst teaching to rules, whereby the answer can 
be obtained. The 'treatment of maxima and minima. Chapter 
III, §§ 2, 3, and curve tracing, Chapter III, § 5 and Chapter VII, 
§ 10, will serve to show what is here meant. 

It is, however, also essential that the student receive thorough 
training in the formal processes and the technique of the Cal- 
culus, and this side has been treated with care and complete- 
ness. Note, for example, the differentiation of composite 
functions in Chapter II, § 8, and the exposition of the use of 
differentials in differentiating in Chapter IV, §§ 4, 5. 

An important application of the graphical inethods, with 
which the Calculus is so intimately associated, is that of 
solving approximately numerical equations which do not 
come under the standard rules of algebra and trigonometry. 
Hitherto, however, little attempt has been made to present 
this subject, simple as it is, in any systematic and elementary 
manner. In Chapter VII the common methods in use by 
physicists and others who apply the Calculus are set forth 
and illustrated by simple examples. 

The book might have included a brief treatment of curva- 
ture and evolutes, and the cycloid. But probably most 



VI PREFACE 

teachers of the Calculus will prefer to take up integration 
next, and so the closing chapter is devoted to the last of the 
elementary functions, the inverse trigonometric functions, with 
special reference to their one great application in the elements 
of mathematics, namely, their application to integration. 

The book is so written that it can be adapted, if desired, to 
an abridged course, in which, after the fundamentals of the 
first three chapters have been covered, any of the remaining 
topics can be treated briefly, and thus a wide scope in subject 
matter is possible, even when the time is short. 

Cambridge, Massacbusbtts, 
January, 1921. 



CONTENTS 



CHAPTER I 
INTRODUCTION 



1. Functions 

2. Continuation. 



General Definition of a Function 



PAGE 
1 

10 



CHAPTER II 

DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. GENERAL 
THEOREMS 



1. Definition of the Derivative . 

2. Differentiation oi x" . 

3. Differentiation of a Constant 

4. Differentiation ot '^x . 

5. Three Theorems about Limits. Infinity 

6. General Formulas of Differentiation 

7. General Formulas of Differentiation, Continued 

8. General Formulas of Differentiation, Concluded 

9. Differentiation of Implicit Algebraic Functions 



13 
16 
20 
21 
22 
29 
32 
35 
39 



CHAPTER III 
APPLICATIONS 

1. Tangents and Normals .46 

2. Maxima and Minima 49 

3. Continuation : Auxiliary Variables 53 

4. Increasing and Decreasing Functions 60 

5. Curve Tracing ... 64 

6. Relative Maxima and Minima. Points of Inflection . . 67 

7. Necessary and Sufficient Conditions 71 

8. Velocity; Rates 72 

vii 



VUl CONTENTS 

CHAPTER IV 
INFINITESIMALS AND DIFFERENTIALS 

PAGE 

1. Infinitesimals ... 81 

2. Continuation. Fundamental Theorem .... 87 

3. Differentials 91 

4. Technique of Differentiation 95 

5. Continuation. Differentiation of Composite Functions . 100 

CHAPTER V 
TRIGONOMETRIC FUNCTIONS 

1. Radian Measure .... .... 105 

2. Differentiation of sin x 110 

3. Certain Limits 112 

4. Critique of the Foregoing Differentiation .... 116 

5. Differentiation of cos x, tan x, etc 117 

6. Shop Work . 118 

7. Maxima and Minima 121 

8. Tangents in Polar Coordinates 128 

9. Differential of Arc 134 

10. Rates and Velocities 138 

CHAPTER VI 
LOGARITHMS AND EXPONENTIALS 

1. Logarithms 146 

2. Differentiation of Logarithms 151 

1 

3. The Limit hm (1+0' 155 

4. The Compound Interest Law 156 

5. Differentiation of e^ 167 

6. Graph of the Function x" 160 

7. The Formulas of Differentiation to Date .... 161 

CHAPTER VII 
APPLICATIONS 

1. The Problem of Niunerical Computation .... 166 

2. Solution of Equations. Known Graphs .... 166 



CONTENTS IX 

PAGE 

3. Interpolation 170 

4. Newton's Method ... .... 172 

5. Direct Use of the Tables 176 

6. Successive Approximations 180 

7. Arrangement of the Numerical Work in Tabular Form . 184 

8. Algebraic Equations 187 

9. Continuation. Cubics and Biquadratics . . 189 
10. Curve Plotting . . 195 

CHAPTER VIII 
THE INVERSE TRIGONOMETRIC FUNCTIONS 

1. Inverse Functions ... ... 206 

2. The Inverse Trigonometric Functions . . • 209 

3. Shop Work 215 

4. Continuation. Numerical Computation .... 218 

5. Applications 221 



CALCULUS 

CHAPTER I 
INTRODUCTION 

The Calexilus was invented in the seventeenth century by 
the mathematician, astronomer, and physicist, Sir Isaac Newton 
in England, and the philosopher Leibniz in Germany. The 
reaction of the invention on geometry and mathematical physics 
was most important. In fact, by far the greatest part of the 
mathematics and the physics of the present day owes its 
existence to this invention. 

1. Functions. The word function, in mathematics, was 
first applied to an expression involving one or more letters 
which represent variable quantities ; as, for example, the 
expressions 

(a) a?, 23?-Z'x + l; 

(6) V«, Va'^ — x^ ; 

,. x^ xy ax + by . 

a + x x^ + y^ Va;2 ^yij^e2 

(d) sin a;, logs;, tan^isc. 

In the second example under (6), two letters enter ; but a 

is thought of as chosen in advance and then held fast, x alone 

being variable. A quantity of this kind is called a constant. 

Thus 

ax + o 

is a function of x which depends on two constants, a and 6. 

1 



2 ' CALCULUS 

Such expressions are written in symbolic, or abbreviated, 
form as f(x), f(x, y) (read : "/ of x," "f of x and y " etc.) ; 
other letters in common use being F, <f>, $, etc.* Thus the 
equation 

(1) f{x) = 2a?-'&x + l 

defines the function /(«) in the present case to be 2a!'— 3a;+l. 
Again, 

(2) <j,{x, y, e) = x"- + y' -\- «• 

is an equation defining the function <j>{x, y, z) as x^ + y^ + z*. 

We shall be concerned for the present with functions of one 
single variable, as illustrated by (1) above. Here, x is called 
the independent variable, since we assign to it any value we 
like. The value of the function, or more briefly, the function, 
is called the dependent variable, and is often denoted by a 
single letter, as ^ =/(«) 

or y = 23? — Zx + l. 



Graphs. A function 
of a single variable, 

y =/(<»), 

can be represented 
geometrically by its 
graph, and this repre- 
sentation is of great aid 
in studying the proper- 
ties of the function. 
The independent vari- 
able is laid off as the 
as-coordinate, or ab- 
scissa, and the depend- 
_ ^ ent variable, or func- 

* To distinguish between /(x) and F(x), read the first " small/ of x " 
and the second, " Uu^e Fotx." 




INTRODUCTION 3 

tion, as the y^oordinate, or ordinate. Thus the graph of the 
function , , 

is the curve 

Illustrations from Geometry and Physics. The familiar 
formulas of geometry and physics afford simple examples of 
functions. Thus the area, A, of a circle is given by the 

formula 

A = ■jtr\ 

where r denotes the radius, n being the fixed number 3.1416. 
Here, r is thought of as the independent variable, — it may 
have any positive value whatever, — and A is the function, or 
dependent variable. 

Again, for the three round bodies, the volumes are : 

(a) F= firr^, sphere ; 

(6) F=irr%, cylinder; 

(c) V='^r%, cone. 

In (6) and (c), h denotes the altitude and r, the radius of 
the base ; V is here a function of the two independent 
variables, r and h. 

The surfaces of these bodies are given by the formulas : 

(a) S = Attt^, sphere ; 

(/8) S = 2Trrh, cylinder ; 

(y) S = tttI, cone ; 

I, in the last formula, denotiag the slant height. Thus we 
have three further examples of functions of one or of two 
variables. 

The formula for a freely falling body is 

s = \gt\ 

where s denotes the distance fallen and t the time ; $> is a 
constant, for it has just one value after the units of time and 



4 CALCULUS 

length liave been chosen. Here, t is the independent variable 
and 8 is the function. If, however, we solve this equation 
for f : , 

then s becomes the radependent variable and t, the function. 
Sometimes two variables are connected by an equation, as 

pv = c, 

where p denotes the pressure of a gas and v its volume, the 
temperature remaining constant. Here, either variable can 
be chosen as the independent variable, and when the equation 
is solved for the other variable, the latter becomes the de- 
pendent variable, or function. Thus, if we write 

c 
P 
p is the independent variable, and v is expressed as a function 

of p. But if we write 

c 
P = -, 

V 

the rSles are reversed. 

The Independent Variable Bestricted. Often the independent 

variable is restricted to a certain interval, as in the case of the 

function 

y = Vo^ — x\ 

Here, x must lie between — a and a : 



■ a<^ x<^a. 






since other values of x make a^ — x^ negative, and the above 
expression has no meaning. 

This was also the case with the geometric examples above 
cited. There, r, h, I were necessarily positive, since there is 
no such thing, for example, as a sphere of zero or negative 
radius. 

The independent variable may also be restricted to being a 
positive whole number, as in the case of the sum of the first n 



INTRODUCTION 

terms of a geometric progression : ' 

s„ = a + ar + ar^ + — + ar'-K 
Here, 

a— ar" 



1 — r 
Suppose a = 1, r = ^, the progression thus becoming 



Then 



1 + 1 + 1+.. .+J-. 
2 22 ^ 2"-i 



-i.-^ 



2 

and we have an example of a function with the iadependent 
variable a natural number, i.e. a positive integer. 

In the case of ihe functions treated in the calculus, the do- 
main of the independent variable is a continuum, i.e., for func- 
tions of a single variable, an interval, as 

a-^x<b, or < as. 

Ordinarily, the later letters of the alphabet, particularly 
X, y, z, are used to represent variables, the early letters denot- 
ing constants. Thus it will be understood, when such an ex- 
pression as 

aad' + OX + C 

is written down, that o, 6, c are constants and x is the variable. 
Multiple-Vahied Functions; Principal Value. The expres- 
sions above cited are all examples of single-valued functions ; 
i.e. to each value of the independent variable x corresponds 
but one value of the function. A function may, however, be 
muUiple-valued ; as in the case of the function y defined by 

the equation 

a,2 + ^2 _ a\ 

Here 

y = ± Va^ — x^, 



6 



CALCULUS 



and so is a double-valued function. This function is, however, 
completely represented by means of the two single-valued 
functions, 

y = ^a^—x^ and y = — Va* — iC*. 



Oraphofy,when 




Graph of 


f 


\. 





x=a 



Fig. 2 



They form the branches of this multiple-valued function. 

The student should notice that the radical sign y' is defined 
as meaniQg the positive square root, not either the positive or 
the negative square root at pleasure. If it is desired to ex- 
press the negative square root, the minus sign must be written 
in front of the radical sign, — -y/. Thus Vi = 2, and not — 2. 
This does not mean that 4 has only one square root. It means 
that the notation V4 calls for the positive, and not for the 
negative, of these t;wo roots. 

Again, 



and not - 2. For (- 
And, generally, 

(1) 



2)2= 4, and y/ means the positive root. 



1 ■\/s^ = — x, 



if X is positive ; 
if X is negative. 



A similar remark applies to the symbol tJ/, which is like- 
wise used to mean the positive 2nth root. Moreover, 



a^=^^, 



'=VE. 



The function 



y = -\/x 



INTRODUCTION 7 

is often called the principal value of the double- valued function 
defined by the equation 

Since multiple-valued functions are studied by means of 
single-valued functions, it -will be understood henceforth, un- 
less the contrary is explicitly , stated, that the word function 
means single-valued function. 

Absolute Value. It is frequently desirable to use merely 
the numerical, or absolute value of a quantity, and to have a 
notation for the same. The notation is: |a!|, read " absolute 
value of X." Thus 

|-3| = 3 and |3|=3. 

We can now write in a single formula what was formerly 
stated by the two equations (1), namely the definition of the 
radical sign, V • 

(2) V^ = |a|. 

Again, by the difference of two numbers we often mean the 
value of the larger less the smaller. Thus the difference of 4 
and 10 is 6 ; and the difference of 10 and 4 is also 6. The 
difference of a and 6, in this sense, can be expressed as either 



Continuous Functions. A function, /(«), is said to be con- 
tinuous if a slight change in x produces but a slight change in 
the value of the function. Thus the polynomials are readUy 
shown to be continuous ; cf. Chap. II, § 5, and all the func- 
tions with which we shall have to deal are continuous, save at 
exceptional points. 

As an example of a function which is discontinuous at a 
certain point may be cited the function (see Tig. 3) 

/(.)=i. 



8 



CALCULUS 



y= 



When X approaches the value 0, the function increases nu- 
merically without limit. The graph of the function has the 
axis of y as an asymptote. 

The fractional rational functions are continuous except at 
the points at which the denominator vanishes. 

1^ Thus the function 

^'-'-^ 

is continuous except at the points 
a; = 1 and a; = — 1. Here, the 
function becomes infinite. Its 
graph is the curve 

_ x'^ + 1 
y~{x-l)Qc + iy 

which evidently has the lines x = l 
and a = — 1 as asymptotes. 
The function 

Fig. 3 /(a;) = tan a; 

is continuous except when x is an odd midtiple of ir/2, 

^ 2w + l 

X ^ -r — TT. 



EXERCISES 

1. If f{x) =x^-4:X + 3, 

show that /(1) = 0, /(2)=-l, /(3) = 0. 
Compute /(0),/(4). Plot the graph of the function. 

2. If " ^(a;) = 4a!' 
compute ^(2) and ^(V3). 

3. If F(x): 



2,x-Z 



x + 7 ' 
compute F{-\/2) correct to three significant figures. 



Ans. -.0204. 



INTRODUCTION 9 

4. If $(«) = (a;' — a;)siii x, 
find all the values of x for -whicli 

*(0)=0. 

5. If ^(a!)=a!^ — a;"^, 
findi/-(8). 

6. Solve the equation 

3^ — xy -\- 3 = By 
for y, thus expressing ^ as a function of x. 

7. If f(x)=ci^, ■ 
show tlmt /(a')/(2/) =/(» + y)- 

8. If y= ^ + ^, 

express a; as a function of y. 

9. Draw the graph of the function 

/(a;)=cB2 + 4a! + 3, 

taking 1 em. as the unit. 

Suggestion : Write the function in the form, (x + l)(a; + 3). 

10. Draw the graph of the function 

f[x) = a^ — 4 a;. 

11. Draw the graph of the function 

and hence illustrate the two discontinuities which this func- 
tion has. 

12. Draw the graph of the function 

f( X 1 1 

^^^' a? {x-lf 



10 CALCULUS 

13. For what values of x are the following functions dis- 
continuous ? 

(a) /(») = cot » ; (c) f(x) = esc a; ; 

(6) f(x) = sec a; ; (d) /(«) = tan |- 

14. Express the double-valued function defined by the 
equation a?-f = -l 

in terms of two single-valued functions. 

15. Express the quadruple-valued function defined by the 
equation y* - 2j/^ + x'' = 

in terms of four single-valued functions. 

16. Express the sum s„ of the first n terms of the arithmetic 
progression 

a+{a + b)+(a + 2b)+ — +(a + n — lb) 

as a function of n. 

Thus obtain the sum of the first n positive integers as a 
function of n. 

17. If P dollars are put at simple interest for one year at 
r per cent, (a) express the amount A (principal and interest) 
as a function of P and r. (b) Express the amount A at the 
end of n years, the interest being compounded annually, as a 
function of P, r, and n. (c) Express the amount A at the 
end of one year, if the interest is compounded m times in the 
year at equal intervals, as a function of P, r, m. 

2. Continuation. General Definition of a Function. The con- 
ception of the function is broader than that of the mathemati- 
cal formulas mentioned in the last paragraph. Let us state 
the definition in its most general form. 

Definition of a Function. 77ie variable y is said to be a 
function of the variable x if there eadsts a law whereby, when x 
is given, y is determined. 



INTRODUCTION 11 

Consider, for example, a quantity of gas confined in a cham- 
ber, — for instance, the charge of the mixture of gasolene and 
air as it is being compressed in the cylinder of an automobLle. 
The charge exerts at each instant a definite pressure, p, of so 
many pounds per square inch on the walls of the chamber, 
and this pressure varies with the volume, v, occupied by the 
charge. In the small fraction of a second under consideration, 
presumably but little heat is gained or lost through the walls 
of the chamber, and thus p is a function of v, 

In this ease, the function is given approximately by the math- 
ematical formula _ 

where C denotes a certain constant. But that which is of first 
importance for our conception is not the formula, but the fact 
that to each value of v there corresponds a definite value of p. 
In other words, there is a definite graph of the relation be- 
tween V and p. The representation of the relation by a math- 
ematical formula is, indeed, important; but what we must 
first see clearly is the fact that there is a definite relation to 
express. 

As another illustration take the curve traced out by the 
pen of a self-registering thermometer of the kind used at a 
meteorological station. The instrument consists of a cylindri- 
cal drum turned 
slowly by clock- 
work at uniform 
speed about a 
vertical axis, a 
sheet of paper ^la. 4 

being wound 

firmly round the drum. A pen is held against the paper, and 
the height of the pen above a certain level is proportional to 
the height of the temperature above the temperature corre- 



12 CALCULUS 

sponding to that level. The apparatus is set in operation, and 
when the drum has been turning for a day, the paper is taken 
off and spread out flat. Thus we have before us the graph of 
the temperature for the day ia question, the independent 
variable being the time (measured in hours from midnight) 
and the dependent variable being the temperature, represented 
by the other coordinate of a point on the curve. 

One more illustration, — that of the resistance of the atmos- 
phere to a rifle bullet. This resistance, measured in pounds, 
depends on the velocity of the bullet, and it is a matter of 
physical experiment to determine the law. But that which 
is of first importance for our conceptions is the fact that there 
is a law, whereby, when the velocity, v, is given an arbitrary 
value withia the limits of the velocities considered, there cor- 
responds to this V a definite value, B, of the resistance. We 
say, then, that B is a, function of v and write 

B = fiv). 

In this connection, ef. the chapter on Mechanics, § 7, Graph 
of the Resistance, La the author's Differential and Integral 
Gdkulus. 



CHAPTER II 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 
GENERAL THEOREMS 

1. Definition of the Derivative. The Calculus deals with 
varying quantity. If j/ is a function of x, then x is thought 
of, not as having one or another special value, but as flowing 
or growing, just as we think of time or of the expanding cir- 
cular ripples made by a stone dropped into a placid pond. 
And y varies with x, sometimes increasing, sometimes decreas- 
ing. Now if we consider the change in x for a short interval, 
say from x=:Xotox= x', the corresponding change in y,a,sy 
goes from y^ to y', will be iu general almost proportional to 
the change in x. For the ratio of these changes is 

x' — Xa 

and this quantity changes only slightly when x' is nearly 
equal to Xq. Let us study this last statement minutely. 




Fig. 5 



The above ratio has a simple geometric meaning, if we draw 
the graph of the function ; for 

PM=x'-Xa; MP' = y'-y^, 

13 



14 CALCULUS 

aj'-OTo 

where t' denotes the angle which the secant PP makes with 
the axis of x. Now let x' approach ojq as its limit. Then r' 
approaches as its limit the angle t which the tangent line of 
the graph at P makes with the axis of x, and hence 

lim y ~yo = tan r 

x'^ x' — !Cd 

( read : "limit, as x' approaches Xg, of ^ ~^° )• 

\ X' — Xf, J 

The determination of this limit and the discussion of its mean- 
ing is the fundamental problem of the Differential Calculus. 

Such are the concepts which underlie the idea of the deriva- 
tive of a function. We turn now to a precise formulation of 
the definition. Let 

(1) y=f('») 

be a given function of x. Let Xq be an arbitrary value of x, 
and let ^o ^ ^^^ corresponding value of the function : 

(2) 2/0 =/(ai)). 

Give to X an Lucrement,* Ax ; i.e. let x have a new value, x', 
and denote the change in x, namely, a/ — aj^, by Aa; : 

a;' — a!o = Aa;, a;' = a^ + Aa;. 

The function, y, will thereby have changed to the value 

(3) 2/' =/(«'') 

and hence have received an iucrement. Ay, where 
y'-yo = Ay, y' = yo + %• 

• The student must not think of this symbol as meaziing A times x. 
We might have used a single letter, as h, to represent the difference in 
question : x'=Xo + h; hut h would not have reminded us that it is the 
increment of x, and not of y, with which we are concerned. The notation 
is read " delta x." 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 15 



Equation (3) is equivalent to the following : 

(4) ' yo + Ay =/(a'o + Aaj). 

From equations (2) and (4) we obtain by subtraction the 
equation 

A2/=/(a!o + Aa;)-/(a!o), 
and hence 

C5-V Ay^ /(a!o+Aa!)-/(a;o) 

^ ^ Ak Aaj 

Definition oi" a Dbeivativb. The limit which the ratio 
(5), namely — ^ , approaches when Ax approaches zero : 

(6) lim^ or li^/^^±Mr^£M, 

tix^Ax AxiO Ax 

is called the derivative of y with respect to x and is denoted by 
Bjy or DJioi) (read : "D x oiy"): 



(J) 



ox^Ax 



In this definition Ax may be negative as well as positive, and 
the limit (6) must be the same when Aa; approaches from the 
negative side as when it approaches from the positive side. 

To differentiate a 
function is to find its 
derivative. 

The geometrical in- 
terpretation of the 
analytical process of 
differentiation is to 
find the slope of the 
graph of the function. 
Por, 

tanr'=^ 



y 


yC^ 


1 




S^' 


Ay 






— ^ ^-ss^^Aa;' 








vi 


y 


X 





aiQ, i 


»' 





Ax 



Fig. 6 



and 



tanr = lim,tanT' =lim— ^ = Z)^. 
F'^P ^x^ Ax 



16 CALCULUS 

2. Differentiation of x". Suppose n has the value 3, so that 
it is required to diEEerentiate the function 

(1) y = a?. 

We must follow the definition of § 1 step by step. Begin, 
then, by assigning to a; a particxilar value, a^, which is to be 
held fast during the, rest of the process, and compute from 
equation (1) the corresponding value yoofy: 

(2) 2/0 = XgK 

Next, give to x an arbitrary increment. Ax, denote the corre- 
sponding increment in y by Ay, and compute it. To this end 
we first write down the equation 

(3) ya + Ay={xo+As:y. 

The right-hand side of this equation can be expanded by the 
binomial theorem, and hence (3) can be written in a new form :* 

(4) yo+Ay = 00o^ + 3 x^^Ax + 3 x^Ax^ -(- Aa?. 

Subtract equation (2) from equation (4) : 

Ay = Sxg^Ax + 3 XgAx'^ + Aa^. 

Next, divide through by Ao; : 

^ = SV + 3xJ^x + Ax''. 
Ax 

We are now ready to let Ax approach as its limit : 

lim M = lim (3 a;„2 + 3 x^Ax + Ax^). 

*It is at this point that the specific properties of the functional come 
into play. Here, it is the'binomial theorem that enables us ultimately 
to compute the limit. In the differentiations of later paragraphs and 
chapters it will always h.e some characteristic property of the function in 
hand which will make possible a transformation at this point. 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 17 

The limit of the left-hand side is, by definition, D,y. On the 

right-hand side, each of the last two terms in the parenthesis 

approaches the limit 0, and so their sum approaches 0, also. 

The first term does not change with Aw. Hence, the whole 

parenthesis approaches the limit Swo^. We have, then, as the 

final result : ^ o a 

■0,2/ = 3 V- 

The subscript has now served its purpose, which was, to 
remind us that Xq is not to vary with Ax, and it may be 
dropped. Thus d^^=3xK 

The differentiation of the function x" in the general case- 
that n is any positive integer can be carried through in pre- 
cisely the same manner. As the result of the first step we have 

(6) ' 2/0 = aso"- 

Next comes : 

(6) y„ + Ay = (Xo + Aa;)», 

and we now apply the binomial theorem to the expression on 
the right-hand side. Thus 

(7) yo + ^y = xo" + mV^Ak + '^^^~J'^ a;o»-2Aa;2 H h Aa;». 

On subtracting (6) from (7) we have : 

Ay = nXo"-^Ax + ^^"' ~J'^ x^^-^-Ax"- H h Aa;". 

1 ■ ^ 

Now divide through by Ax : 

^ = nxo"-^ + '^^"'~'^\ --^Ax +■■■■+ Ax--^ 
Ax 1 '2 

and let Ax approach the limit zero : 

lim ^ = lim f «a!o"-i + "^" ~ ^^ V'^Aw -H ■ ■ • -I- Aa!»-i\ 

Arm Ax /^asiO \ 1-2 J 

Each term of the parenthesis after the first is the product 
of a constant factor and a positive power of Aa;. This second 



18 CALCULUS 

factor approaches zero when Aa; approaches zero ; consequently 
the whole term approaches zero. There is only a fixed num- 
ber of these terms, and so the whole parenthesis approaches 
the limit waso""". Hence 

On dropping the subscript we obtain the final result : 

(8) Z),£B» = raa!»-i. 
In particular, if n = 1, we have 

(9) D,x = l. 

EXERCISES 

Differentiate the following seven functions, applying the 
process of § 1 step by step. , 



1. 


y = 4.a?. 


Ans. Dji = 12x\ 


2. 


y = a^. 




3. 


y = 2f -3a;+.l. 


Ans. Djy = 4 a; — 3. 


4. 


y = x' — a?. 


Ans. D^ = 7afi-5!ei. 


5.* 


/(a!)=l-2a^. 


Ans. DJ(x)=-8a?. 


6. 


,i>{x)=x-'--2x + l. 




7. 


F{x) = {l-xy. 




8. Let 


y = 5x- x\ 





and take a;o = 1 ; then yg = 4. If Aa; = .2, then Aj/ = .56 and 

^ = 2.8. Show further that, 
Aa; 

for Aa;=.l, A;/ = .29, ^=2.9; 

Aa; 
and . 

for Aa; = .01, Ay = .0299, ^=2.99. 

Aa; 

* It is immaterial wliethier we write 

y = l-2x* or f(x) =1-2**. 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 19 



Plot the curve accurately for values of x from a; = to 
x = 5, taking 1 cm. as the unit, and draw the secants* in 
each of the three foregoing cases. 

What appears to be the slope of the curve at the point 
{xq, yo) = (l> 4)? Prove your guess to be correct. 

9. In Ex. 7, let a!o = — 1- H Ax is given successively the 

values .01 and — .01, compute Ay and ^• 

Ax 

10. Complete the following table : 



Ax 


Ay 


tanr' = ^ 
. Ax 


.1 

.01 

.001 







for each of the functions : 

(a) y = x^-2x + l, «„ = 2 ; 

(&) y = x — a?, iCi, = - 1 ; 

(c) y = 3ce2 — x, cc,, = 0. 

11. By means of the general theorem (8) write down the 
derivatives of the following functions : 

a^; a^; a!""; x; aP. 

By means of the definition of § 1 differentiate each of the 
following functions : 

Ans. D^= -. 



12. 

13. 
14. 



1 

y=— 

X 

1 



y = 



Q? 



Ans. Dji = — 
Ans. Dji = — 



* The student shoidd recall from his earlier work how to draw a straight 
line on squared paper when a point and the slope ot the line are given. 



20 CALCULUS 

3. Derivative of a Constant. The function 

where c denotes a constant, has for its graph a right line paral- 
lel to the axis of x. Since the derivative of a function is repre- 
sented geometrically by the slope of its graph, it is clear that 
the derivative of this function is zero : 

Dji = 0. 

It is instructive, however, to obtain this result analytically 
by the process of § 1. We have here : 

2/o=/(9i))=c, 
yo + Ay =/(a!o4- Aa;)= c; 

hence Aw = and —2 = 0. 

" Aa; 

Now allow Aa; to approach 0. The value of Ay/Aa; is always 
0, and hence its limit * is : 

lim^=0, or Z),c = 0. 

Ax^ Aa; 

* We note here an error frequently made in presenting the subject of 
limits in school mathematics. It is there often stated that " a variahle JT 
approaches a limit Ait X comes indefinitely near to A, but never reaches 
A." This last requirement is not a part of the conception of a variable's 
approaching a limit. It is true that it is often inexpedient to allow the 
independent variable to reach its limit. Thus, in differentiating a func- 
tion, the ratio Ay/ Ax ceases to have a meaning when As = 0, since divi- 
sion by is impossible. The problem of differentiation is not to find the 
value of Ay I Ax, when Ax = ; such a question would be absurd. What 
we do is to allow Aa; to approach zero as its limit without ever reaching 
that limit. We can do this for the reason that Az is the iTidepemdent 
«ario6ie. 

When, however, it is Ay or Ay /Ax that is under consideration, we have 
to do with dependevi ■variables, and we have no control over them, as to 
whether they reach their limit or not. Thus in the case of the text both 
Ay and Ay/ Ax are constants (=0). When Aa; approaches 0, they always 
have one and the same value, and so, under the correct concei>tion of 
approach to a limit each approaches a limit, namely 0. 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 21 

We can state the result by saying : The derivative of a con- 
stant is 0. 

4. Differentiation of Va;. Let us differentiate 
y=-\/x. 
Here, 2/0 = V«0) 2/o + % = Va;o + Aa;, 



Ay _ Vgp + Aa; — VajQ 
Ax Aa; 

We cannot as yet see what limit the right-hand side approaches 
when Aa; approaches 0, for both numerator and denominator 

approach 0, and - has no meaning. We can, however, trans- 
form the fraction by multiplying numerator and denominator 
by the sum of the radicals and recalling the formula of Elemen- 
tary Algebra : „= _ 52 =(„ _ 6)(„ + j). 



Thus % — V^'o + Aa; — Vajp _ Vxg + Aa; -j- Vxp 
^a; Aa; Va;o + Aa!+V% 

_ 1 (xg + Aa;) — a;o 1 

^^ Vxo + AX + Vwo -Vxq + Aa; -|-' V^' 

and hence lim — ^ = lim — ^ ^ = ^^ • 

a^ Ax i»^ Va!o -(- Aa; -H Va;o 2 VaJo 

Dropping the subscript, we have : 



2Vm 

EXERCISES 

1. Differentiate the function'^/ = Ans. Dj) = — 



Vx 2-\/a? 

2. If 2/=V2-3a;, 

3 

show that Z)j« = =^=r ■ 

2V2-3a; 



22 CALCULUS 



3. Prove: D,Vl — x = — 



4. Prove: 2)^Va + 6a; = 



2Vl-a! 
6 



2 Va + bx 



5. Three Theorems about Limits. Infinity.* In the further 
treatment of differentiation the following theorems are needed. 

Theobem I. The limit of the sum of two variables is equal to 
the sum of their limits: 

Urn {X+Y)= lim X + lim T. 

In this theorem we think of X and T as two dependent 
variables, each of which approaches a limit : 

limX=^, limr=JB. 

We do hot care what the independent variable may be. In 
the applications of the theorem to computing derivatives, the 
independent variable will always be Ax, and it will be allowed 
to approach 0, without ever reaching its limit. 

Since X approaches A, it comes nearer and nearer to this 
value. Let the difference between the variable and its limit 
be denoted by £ ; then the limit of e is : 



(1) 




X- 


-A-. 


= ^, 


X = 


= A + €; 


Similarly, 


let 






lime 


= 0. 




(2) 




Y- 


■B = 


=,'?> 


T= 


■■B + r,; 


then 








limr 


, = 0. 





* This paragraph should he read carefully and its content grasped, 
but the student should not be required to reproduce it at this stage of his 
work. He will meet frequent applications of its principles, and he should 
turn back each time to these pages and read anew the theorem involved, 
with its proof. When he has thus come to see the full meaning and im- 
portance of these theorems, he should demand of himself that he be able 
readily to reproduce the proofs. 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 23 

It will be convenient to think of these numbers as repre- 
sented geometrically by points on the scale of numbers, thus : 

— t — 1 nt — r-t- 

Fig. 7 

Of course, A and B may be negative or 0. e and rj may be 
negative as well as positive, or even 0. 

Consider the variable X+Y. Its value from (1) and (2) is : 

X + T=A + B + € + r,. ^„, , ,. y 

Hence lim (X+Y) =lim(A + B + e + rj). 

But since lim e = and lim 17 = 0, the limit of the right-hand ^ 
side of this equation is A + B, or 

lim{X + Y)=A + B. 

Consequently, lim (X+Y)= lim X + lim P, q. e. d. 

CoBOLLAKY. The limit of the sum of any fined number of 
variables is equal to the sum, of the limits of these variables : 

lim (Xi + X2+-+X,)= lim Xi + lim Xj -1- - -1- lim X„. 
Suppose n = 3. Then 

Xi + X2 -I- X3 = (Xi -f- X2) -f X3. 
From Theorem I it follows that 

lim (Xi -f- X2 + X3) = lim (Xi -|- X2) -1- lim Xs- 
Applying the Theorem again, we have 

lim (Xi + Xj) = lim Xj -|- lim Xj. 

Hence the corollary is true for n = 3. It can how be estab- 
lished for ji = 4 ; and so on. By the method of Mathematical 
Induction it can be proven generally. Or, the proof of the 
main theorem may be extended directly to the present 
theorem. 



24 CALCULUS 

Theorem II. Tlie limit of the product of two variables is 
equal to the product of their limits : 

lim (XF) = (limX)(lim T). 
From equations (1) and (2) it follows that 

or XT= AB + B€ + Ari + erj. 

Hence lim 'XT= lim (AB + B€ + A7i + eq). 

Since A and B are constants, each of the last three terms in 
the parenthesis approaches the limit 0, and so the limit of the 
parenthesis is AB. Hence 

lim{XY)= AB, 

or lim {XT) = (lim :X)(lim T), q. e. d. 

Corollary. The limit of the product of n variables is equal 
to the product of the limits of these variables : 

lim (XiXj ... X„) = (lim Xi)(lim X^) - (limX„). 

The proof is similar to that of the corollary under Theorem I. 

Remark. As a particular case under Theorem II we have : 
lim(C'X) = C(limX), 
where O is a constant. 

Theobbm III. The limit of the quotient of two variables is 
equal to the quotient of their limits, provided that the limit of the 
divisor is not ; ^ , . ^ 

From equations (1) and (2) above we have : 

X^ A + t _ 
Y B + 7, 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 25 

Subtract AfB from each side of this equation and reduce : 

X A_ A-^€ A_ Bt — Ar] 
Y B B + 7] B B' + Br,' 



Hence 


X 
T 


■4- 


B€ — Ari 

B' + 5,' 


and 


lim^=lim(^- 
Y \B 


Be- J 

B^ + B 


We wish to show that 








lim 


Bc- 


1^ = 0. 



£2 + Br] 

The numerator is seen at once to approach zero. The limit of 
the denominator is &. Let fl^ be a positive number less than 



H B' 

Fig. 8 

B\ Then the denominator will finajly become and remaiu 
greater than H, and hence the numerical value of the quotient 
in question will not exceed the numerical value of 

■Be- J.T; 



H 

But the limit of this expression is zero, and hence 

,. X A 
Y B 

,. X limX „ „ ^ 

or lim — = - — — , q. e. d. 

Y lim Y 

In particular, we see that, if a variable approaches unity as 
its limit, its reciprocal also approaches unity : 

If limX=l, then lim^ = l. 



26 CALCULUS 

where (7 is a constant and lim X^O. 

Remark. If the denominator Y approaches as its limit, 

no general inference about the limit of the fraction can be 

drawn, as the following examples show. Let T have the 

values : 

F=i J- -J- -^ 
10' 100' 1000' ' 10"' 

(1) If the corresponding values of X are : 



102' 1002' 10002' ' 10""' 

X 1 

then lim — - = lim — ■ = 0. 

T 10- 

111 



(2) If X = 



Vip' vioo' viooo' ■' ^^' 



then X/F=10"/^ approaches no limit, but increases beyond 
all limit. 

(3) If X = ^, 



10' 100' 1000' ' 10"' ' 
where c is any arbitrarily chosen fixed number, then 

lim — = c. 
Y 

fA\ Tf "V^ J- 1 1 1 

*■ '' ~io' ~ioo' iooo' ""10,000' ■"' 

then X/ Y assumes alternately the values + 1 and — 1, and 
hence, although remaining finite, approaches no limit. 

To sum up, then, we see that when X and Y both approach 
as their limit, their ratio may approach any limit whatever, 
or it may increase beyond all limit, or finally, although remain- 



DIFFERENTIATION OP ALGEBRAIC FUNCTIONS 27 

ing finite, i.e. always lying between two fixed numbers, no mat- 
ter how widely tlie latter may differ from each other in value, 
— it may jump about and so fail to approach a limit. 

Infinity. If limX=^:?t=0 and limF=0, then X/ Fin- 
creases beyond all limit, or becomes infinite. A variable Z is 
said to become infinite when it ultimately becomes and re- 
mains greater numerically than any preassigned quantity, how- 
ever large.* If it takes on only positive values, it becomes 
positively infinite; if only negative values, it becomes negatively 
infinite. We express its behavior by the notation : 

limZ = oio or lim^ = -|-oo or limZ = — oo. 

But this notation does not imply that infinity is a limit ; the 
variable in this case approaches no limit. And so the notation 
should not be read " Z approaches infinity" or "Z equals 
infinity ; but " Z becomes infinite.'' 

Thus if the graph of a function has its tangent at a certain 
point parallel to the axis of ordinates, we shall have for that 
point : , 

lim — ^ = 00 ; 

read : " Ay /Ax becomes infinite when Ax approaches 0." 

Some writers find it convenient to use the expression "a 
variable approaches a limit " to include the case that the vari- 
able becomes infinite. We shall not adopt this mode of ex- 
pression, but shall understand the words " approaches a limit " 
in their strict sense. 

'*T!f a function f(x) becomes infinite when x approaches a cer- 
tain value a, as for example 

/(a;)=- for a = 0, 

* Note that the statement sometimes made that " Z becomes greater 
than any assignable quantity" is absurd. There is no quantity that is 
greater than any assignable quantity. 



28 CALCULUS 

we denote this by writing 

/(a)=oo 

(or /(a) =+00 or =— oo, if this happens to be the case 
and we wish to call attention to the fact). 
It is in this sense that the equation 

tan 90° =00 

is to be understood in Trigonometry. The equation does not 
mean that 90° has a tangent and that the vcdue of the latter is 
so. It means that, as x approaches 90° as its limit, tana; 
exceeds numerically any number one may name in advance, 
and stays above this number as x continues to approach 90° 
without ever reaching its limit, 90°. 

Definition of a Continuous Function. We can now make more 
explicit the definition given iu Chapter I by saying: f(x) is 
continuous at the poiut a; = a if 

lim/(a!)=/(a). 

From Exercises 1-3 below it follows that the polynomials 
are continuous for all values of x, and that the fractional 
rational functions are continuous except when the denominator 
va.nishes. 

EXERCISES 

1. Show that, if n is any positive integer, 

lim(Z») = (limX)». 

2. If O (x) = Co + C^X + C2X^ -\ + cjif, 

then lim Q{x)= G(a)= (^ + CiU + ofl,^ + - + cjn". 

x=a 

3. If Q{x) and F(x) are any two polynomials and if F{a)^0, 
then lim^M = ^(^. 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 29 

4. If X remains finite and Y" approaches as its limit, show 
^^"-^ lim(Xr)=0. 

5. Show that ^^^ a^ + 1 _ 1 

x™3a;2 + 2a;— 1~3" 

Suggestion. Begin by dividing the numerator and the de- 
nominator by x^. 

Evaluate the following limits : 

6. lim ^ + W-'^^, 7. Km 12^!+5 

;=»a!3-7a; + 3 ^•^''< .=« 4:afi + 3x*-+7 x^-1 

8. lim?^^±^'.„/./. 9. lim^^±^. ^^ 

x=a cx + da;"' ''' *-«^, x^ ex + dx~^ -^T 

A- 



10. lim ■ \ 11. lim — ' ^ 

*-" X »=" VS + 5a;2-|- 4»^ 

12. lim ^^ + ^' + '^ - 13. lim- 



X X=o<, VI + a* 

6. General Formulas of differentiation. 

Theorem I. The derivative of the product of a constant and 
a function is equal to the product of the constant into the deriva- 
tive of the function : 



(I) 


DJ^cu) = cDj.i. 


Por, let 


y = cu. 


Then 


2/0 = CMo, 




yo + Ay = c{ua + Am), 


hence 


Ay = cAm, 




Ay _ Am 
Aa; Aa;' 


and 


lim^ = limfc^\ 
Aa=M Aa; i»io\_ Ax J 



30 CALCULUS 

The limit of the left-hand side is D^. On the right, Aw/Aa; 
approaches D^ as its limit. Hence by § 5, Theorem II, the 
limit of the right-hand side is cD.tt, and we have 

X>,(cm)=cD,m, q.e.d. 

Theorem II. Tim derivative of the sum of two functions is 
equal to the sum of their derivatives : 



(H) 


D^{u + v)=D,u + D,v. • 


For, let 


y = u + v. 


Then 


yo=uo + vo, 




yo + Ay = Mo + Am + -Wo + Atj, 


hence 


Ay = Am + Av, 


and 


Ay _ Am Av 
Ax Ax Ax 



When Ax approaches 0, the first term on the right approaches 
Dji and the second D^v. Hence by § 5, Theorem I, the whole 
right-hand side approaches D^u + D^v, and we have 

lim^ = lim('^ + ^Vl™ — + lim^, 
is.,c^Ax Ai=o\Aa; Ax J ^x^Ax ^x^Ax 

or D^ = D,u + D^v, q. e. d. 

CoROLLAKT. 2%e derivative of the sum of any number of 
functions is equal to the sum of their derivatives. 

If we have the sum of three functions, we can write 
u + V + w = u -\-(v + w). 
Hence -D,(m + v+w)= JD^u + Z>,(v + w) 

= D,u + D,v -t- D^w. 

Next, we can consider the sum of four functions, and so on. 
Or we can extend the proof of Theorem II immediately to the 
sum of n functions. 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 31 

■ Polynomials. We are now in a position to differentiate any 
polynomial. For example: 

= 7D,3^ - 5Dji? + 1 = 280)3 - 153? + 1. 

EXERCISES 

Differentiate the following functions : 

1. y = 2x^ — Zx + l. Ans. D^y = ix — 3. 

2. y=a + bx+ cx\ Ans. D^ —b + 2cx. 

3. y = x^-3<jfi + x — l. Ans. D^ = ia^ — 9 x'- + 1. 

4. y = a + bx + CO? + dsfi. 

5. y = — — Ans. 3a:^ — 6a!' — 1. 

^ 2 

6. jr^. ax--^2hx + c ^^^ ax + b 

2h h 

7. ■n-!B*-3|a;2+V3. Ans. inafi-T^x. 

8. Differentiate 

(a) Vot — 16^2 with, respect to t ; 

(6) a + bs+ cs^ with respect to s ; 

(c) .Qlly^ — 8.15m2/^ — .'dim with respect to y. 

9. Find the slope of the curve 

42/ = a;^ — Bk — 1 
at the point (1, — 2). Ans. — 1. 

10. At what angle does the curve 
^y = AiX — a? 
cut the negative axis of a; ? 



32 CALCULUS 

11. At wtat angles do the curves y = 3? and y = oi? intersect ? 

Ans. 0° and 8° 7'- 

12. At what angles do the curves y = a? — Zx and y = x 
intersect? Ans. ges-BC and 38° 40'. 

7. General Formulas of Differentiation, Continued. 

Theorem III. The derivative of a product is given by the 
formula : 

(III) D^uv) = uD^v + vD^u. 

Let 2/ =^ Mv. r 

Then 2/o = Mo^) 

2/0 + % = (mo + ^«) (''o + ^i>), 
Ay = UyAv + VqAu + AmAv, 

Av Av , Am , . Au 

-^ = Wo 1- ""o 1- Am — . 

Ax Aa; Aa; Ax 

and, by Theorem I, § 5 : 

lim— ^= lim( ?to — |+ Hmf -Wo — )+ lim( Am — )• 
^lioAa; Ax^\ Ax J i^3:^\ Ax J Axiio\ Ax J 

By Theorem II, § 5, the last limit has the value 0, since 
lim Am = and lim (Av/Asc) = Dj). The first two limits have 
the values UgD^v and VgD^u respectively.* Hence, dropping 
the subscripts, we have : 

D^y = uD^v + vDji, q. e. d. 

By a repeated application of this theorem the product of 
any number of functions can be differentiated. When more 

* More strictly, the notation sliould read liere, before the subscripts 
are dropped : [DiO]js=5«||, etc. Similarly in the proofs of Theorem I, II, 
andV. 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 33 

than two' factors are present, the formula is conveniently writ- 
ten in the form : 

(1) D^(uvw) _Dji ^ D^v ^ Djjo 

uvw u V w 

For a reason that will appear later, this is called the loga- 
rithmic derivative of uvw. 



Theokem IV. The derivative of a quotient is given by the 
formula :* 



(IV) 


-0 


_ vD^u — uDj) 

y2 


Let 




u 

y = — 

V 


Then 


^^0=^ 


y, + ^y = '^ 



Vq Vq + Av 

. _ Mo + Am _ Mo _ 'WqAm — u^^v 
Vo + AlJ t'o Va(ya + £^v) ' 

Am Aw 
Aj/ Az Ax 



Ax Vq{vo + Av) 
By Theorem III of § 5 we have : 

limfvo^-uo^) 
/^x^Ax iim[vo(wo + ^«)] " 

Applying Theorems I and II of § 5 and dropping the sub- 
scripts we obtain : 

n „, vDm — uD.v , 

D^y = — ^ — I ^, q. e. d. 

r 

* The student may find it convenient to remember this formula by 
putting it into words: "The denominator into the derivative of the 
numerator, minus the numerator into the derivative of the denominator, 
over the square of the denominator." 



34 CALCULUS 

Example 1. Let _ 2 — 3a; 

^~l-2a!' 

Then i}y- a-2x)DJ2-3x)-(2-3x)DAl-2x) 

'^ (1 - 2xy 

^ (l-2a!)(-3)-(2-3a!)(-2) ^ 1 

(l-2a;)2 (l-2xy' 

Example 2. To prove that the theorem 
X>j.a;" = ma;""' 
is true when n. is a negative integer, n = — m. Here 

„ 1 

a!" = — 

a;" 

TT n „ a!"i)^l — liJjB" wia!"""i _„_1 

Hence Djc" = ^-— = = — nuc "" \ 

On replacing m in this last expression by its value, — n, the 
proof is complete. 

EXERCISES 
Differentiate the following functions : 

-^.^ -v '^ " 1 + x^ '^ 0- + x^y 

■4- ^y^ ' 

_- v+ ' „ a^ ^ r. 3«2 — 2«' 

1 — a; (1 — a!)2 

5. s = Ans. Dfi = ■ ~ 



1 + t ' ' (i + ty 

t±^. Ans ^^ + 208-0" 

» + a ' f? + 2az-ira?' 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 35 
2 ay ax+b 



9. ■:lslji.. 10. 



a2-2/2 


a^ + a^ 


x + a 


a + bx + ciB' 


X 


3 — 405 + a;3 ' 



a? +px + q 



j2 "-^-"-r-^ . 13 ^±1. 

a!2 a? 

8. General rormulas of Differentiation, Concluded. Oomr 
posite Functions. 

Theobbm V. If u is expressed as a function of y and y m 
turn as a Junction ofx: 

u=fQ/), y = 4,{x), 

then 

(V) D^u = D^u-D,y. 

Here ya = 4> ipo), "o =/(2/o) , 

yo + l^y = <j>(xo + A*), Ug + ^u =f(2lo + ^V), 

A«=/(2/o + A2/)-/0/o), 

^M - /(yo + %) -/(yo) . :^. 
Ax Aj/ Aa; 

When Aa; approaches 0, A?/ also approaches 0, and hence the 
limit of the right-hand side is 

Ai^ /(yo + Ay)-/(yo)Y ^^ ^] = DJ{y)D^y. 
\/!lv^ Ay J\^^ AxJ 

The limit of the left-hand side is Dji. Consequently 

Dji = D^u-Dj/, q.e.d. 

This equation can also be written in the form : 
(V) Dj^ = DJ{y)DM«'). 



36 CALCULUS 

The truth of the theorem does not depend on the particular 
letters by •vrhich the variables are denoted. We may replace, 
for example, xhj t and yhj x: 

DfU = D^u D,x. 

Dividing through by the second factor on the right, we thus 
obtain the formula : 

(V") A« = §^- 

D,x 

Example 1. In § 4 we differentiated the function Va;, and 
we saw that other radicals can be differentiated in a similar 
manner. But each new differentiation required the evaluation 
of lim Ay/Aa; by working through the details of a limiting 
process. Theorem V enables us to avoid such computations, 
as the following example will show. 

To differentiate the function 



u = Va^ — «'. 

Let y = a^ — x*. 

Then u = Vy, 

and the differentiation thus comes directly under Theorem V, 
if we set _ 

■/(y) — ^y> ^ (»') = a^ — a;^- 

Hence we have : 

(1) D,u = D^y/y D^a^ - x^). 

Now, the formula jy -^~_ 1 

does not mean that the independent variable must be denoted 
by the letter x. If the independent variable is y, the formula 

reads : _ ^ 

D,-s/~y = -^- 

2V^ 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 37 

Consequently (1) can be written in the form : 
(2) D^ri = -!-(- 2 x) = - "^ 



2V2/ V02- 

We have, then, as the final result : 



D,-\/a? — x^ = - 



Va^ — x'^ 
Exam/pie 2. To differentiate the function 

^ (1 - x/ 
Let z=:l — X. 

Then y = z~K 

To apply Theorem V in the present case, the letters u and y 
must be replaced respectively by y and z. Thus Theorem V 
reads here: D^y = D^Dj^, 

or Dj) = D^^D^a-x). 

Since Formula (8) of § 2 has been extended to negative in- 
tegral values of n by § 7, Ex. 2, we have : 

Z)^==-3«-*. 

Hence D,y = -Zz-\-l)=-. 

z 

or D, 



= (l-a!)8 (1-a;)* 



•x-' 



EXERCISES 



Diiferentia|e A^^lowing functions : 



.-• V 



1. (f'i ' v=Va2 + a;''. ^ras, 



a; 



P> Va2 + a;2 



2.* 2/ = — ■ .dns. 



a; 



Va2 — a;2 V(a2 — a;^)^ 

* Note that Formula (8) of § 2 has also been shown to hold for the 
case n =— ^ ; § 4, Ex. 1. 



38 / CALCULUS 

3. 2/=Vl + a! + ai2. Ans. — "^ "^ 

2 VI + iB 4- «!« 

A 1 < A 1 — 2a; 

4. y = , . Ans. 



V3-2a; + 4a!2 VS — 2 a; + 4 a;^ 

(1 - a;)' (1 - x)* 



~ x^ + 1 .6 + 4a! 

6. u — — • Ans. 



{2-3a!)2 (2-39!)' 

^ (l+2a!)* ^ (2 + a;)3 

9. y=(-^Y. 10. M = ; "^ 



-(r^T 



1 — 2a; + ai2 
11.* M = a;(l — a;)<. Ans. (1 — 5a!)(l — a!)3. 

12. M = a;(a + te)". -4ns. [a + (w + l)6a;](a + 6a;)"~'- 

13. u = x\a + 6a;)". 14. u = a;'(l — x)*. 

15. M = a;Va — a;. ^ns. ~ ^ ■ 

2 Va — X 



16. u = 7?^(^ — ^. 17. M = a;Vl+a; + a!2. 

18. M= "^ - 19. « = - "^ 



Va" - a;2 Vl + a; + a;^ 

20. «=JM^. 21.t ^*=^ i 

^^c + d* (a2-2aa;)< 

22. M = r — -• 23. M = - 



(a;2 — 1)2 1 + a; + a;' 

a;*— 36a;2+362a!-6» „_ a + 6 

24. M = ^ ■ • 25. M = 



■2 



6 — a; {a-irhxY 

* Use Theorem HI. i 

t Do not use Theorem IV. 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 39 

9. Differentiation of Implicit Algebraic Functions. Wlien 
X and y are connected by such a relation as 

a;2 + y2 = a% 

or a^ — 2xy + 2/^ = 0, 

or xysaiy = x + y log x, 

i.e. if y is given as a function of x by an equation, 

F{x, y)=0 or $ (a;, y) = * (a;, y), 

which must first be solved for y, then y is said to be an implicit 
function of x. If we solve the equation for y, thus obtaining 
the equation 

y =/(»), 

2/ thereby becomes an explicit function of x. 

By an algebraic function of a; is meant a function y which 
satisfies an equation of the form 

O{x,y)=0, 

■where G(x, y) is an irreducible polynomial in x and y; i.e. a 
polynomial that cannot be factored and written as the product 
of two polynomials. 

Thus the polynomials are algebraic functions ; for if 

y = ao + aiX+ ■•• + «„»" = P{vi), 

then y satisfies the algebraic equation 

G{x,y)=y-P{x)=0. 

Similarly, the fractions in x are algebraic functions ; for if 

P(xy 



w^here -P(a;) and Q(<c) are polynomials having no common 
factor, then y satisfies the algebraic equation 

G{x,y) = Qix)y-Fix)=.0. 



40 CALCULUS 

The polynomials and the fractions are also called rationed 

functions. Thus, ^ 

■' ax + by 

a;2 + y^ 

is a rational function of the two independent variables x and y. 
Agaia, all roots of polynomials, as 

y = Vl + X + sfi, 
or such functions as 



^1 — X 



are algebraic, as is seen on freeing the equation from radicals 
and transposing. The converse, however, — namely, that every 
algebraic function can be expressed by means of rational func- 
tions and radicals, — is not true. 

In order to differentiate an algebraic function, it is sufB.cient 
to differentiate the equation as it stands. Thus if 

(1) x^ + y^ = a?, 
we have 

(2) Djc''- + Dy = D,a?. 

To find the value of the second term, apply Theorem V, § 8. 
Thus Z>^2 = D^'D^ = 2yDj/. 

This last factor, D^, is precisely the derivative we wish to 
find, and it is given by completiug the differentiations indi- 
cated in (2) : o , o t^ n 

and solving this equation for Djy : 

y 

The final result is, of course, the same as if we had solved 

equation (1) for y : , 

^ y= ± Vo* - x\ 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 41 
and then differentiated : 

Va2 — a? y 

In the case, however, of the equation 

(3) a?-2xy + y^ = 0, 

we cannot solve for y and obtain an explicit function expressed 
in terms of radicals. Nevertheless, the equation defines y as 
a perfectly definite function of x ; for, on giving to x any 
special numerical value, as x = 2, we have an algebraic func- 
tion for y, — here, ^ . „ „ 

and the roots of this equation can be computed to any degree 
of precision. 

To find the derivative of this function, differentiate equation 

(3) as it stands with respect to x : 

(4) Dj>?-2DXxy) + D^y^ = 0. 

The second term in this last equation can be evaluated by 

Theorem III of § 7 : „ , , 

I>,(xy) = xD^y + y, 

where Dj^ denotes the derivative we wish to find. 

To the evaluation of the third term in (4) Theorem V of § 8 
applies: Dy = 5y^Dj,. 

^^^'^^ 3a!2 - 2 xDj^ -2y + 5y*D,y = 0. 

Solving this equation for Dj/, we have as the final result : 

'^ 5y*-2x 

Thus, for example, the curve is seen to go through the point 
(1, 1), and its slope there is 

(-D^)a,i)=-i- 



42 CALCULUS 

The differentiation of implicit functions as set forth in the 
above examples is based on the assumptions a) that the given 
equation defines y as a function of a; ; 6) that this function has 
a derivative. The proof of these assumptions belongs to a 
more advanced stage of analysis. In the case, however, of the 
equations we meet in practice, — for example, such equations 
as come from a problem ia geometry or physics, — the condi- 
tions for the existence of a solution and of its derivative are 
fulfilled, and we shall take it for granted henceforth that this 
is true of the implicit functions we meet. 

Derivative of x", n Fractional. We are now in a position to 

prove the theorem „ , 

Djc" = nx"''- 

for the case that m is a fra,ction. Let 

q 

where p, q are whole numbers which are prime to each other. 
Let , 

y = »'.., 
Then y = x". 

Differentiating each side of this equatiqn with respect to x, 

we have : ^ ^ 

D,r = Djx", 

and since, by Theorem V, § 8, 

D^ = Dj>Djj = qy-^Dj), 

it follows that , _ 

qy-Wjj=px!^\ 

or X>^ = £-— • 

qy 1 

This last denominator has the value 
(a?Y-^ = x '. 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 43 



Hence 
We see, 


then, 


that 
















Ds = 




■.naf' 


•1 
? 



q. e. d. 



If, finally, n is a, negative fraction, n = — m, the proof can 
be given precisely as was done in § 7, Ex. 2. Thus the theorem 

is now established for all commensurable values of n. 

The theorem is true even when n is irrational, e.g. m = jr or 
V2; the proof depends on the logarithmic function and will 
be given when that function has been differentiated. 

Example. Differentiate the function 

y = Va^ — a?. 

' Apply Theorem V, § 8, setting 

z = a^ — x^. 

Then Z>^2/ = !>.»*= D^^D^a = |a"*(- Sk^). 



Hence .D^-\/a? — a^ : 



V(a' - 3?f 



EXERCISES 

1. If 2!B'-3a!2t/ + 4a!2/ + 62/' = 0, 

findD.2/. Ans. j y ^ 6a;^ - 6a^ + 4y 

^ '^ 3a;2 + 4a! + 18yJ 

2. If 2/* — 2 xy'^ = a^, 
find D^. 

3. Show that the curve 

x'^ — 2xy'^+'f + Zx — Sy = 
cuts the axis of x at the origin at an angle of 46°. 



44 CALCULUS 

4. Plot the curve a^ + ^ = 81, 

taking 1 cm. as the unit. Show that this curve is cut orthog- 
onally by the bisectors of the angles made by the coordinate 
axes. 



Differentiate the following functions : 



5. M = Vl — X. Ans. 



1 



6. u = 'Va'' — 2ax + x\ Ans. 



7. M = Vc*— 3c2a!-|-3ca;2- a^. Ans. — ■ 



5 V(l - xy 
-2 
3-s/a-x 
3 



5-Vc^ — 2cx + x^ 

u = \ "^ ■ 9. u = x-i/a + bx + (x>?. 

M — a; 



10. M = 



..<l^^±^. Ans. 



3 Va!2(l - xy 



J J Va-gj+Va + a; ^^^ g^ + aVa^-a;" ^ 

' a!2 Va^ — ofi 

13. r=Va6. 

lO^a;!' 
16. M = a; V2 a;. 





V a — a; — V a + a; 


12. 


yrrz-v'aS^. 


14. 


l-a,-i 
u= ■ 

x^ 


15. 


y= Z- ■ 

■VX 


17. 


■Vt 


18. 


(i/^ + l)^f-y. 


19. 


(«" - o2)^ 



Ans. -2 2 . 

2-Vf^ 

. 3aVs2 — o2 

8* 



DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 45 



20. "-'^ ■ 21. -''' ^-=^ 



V2aa!-a!2 ^ (1 + «)' 



22. « = iB(a2 — x^y. 23. M = (6 — «) V& + «. 

24. !Pind the slope of the curve y = x^ in the point whose 
abscissa is 2. Ans. tan t = .116. 

25. If pv^* = c, find D^p. 

26. If2/V5 = l+a;, find D^j/- ^ns. ^-=4- 

2a;Va! 
.27. Differentiate y in two ways, where 

xy + A:y = Zx, 

and show that the results agree. - ' 

28. The same, when 2/^ = 2 mx. 

29. Show that the curves 

32/ = 2a; + a;y, 2?/ + Sx + j/^ = aj'j/, 

intersect at right angles at the origin. 

30. Find the angle at which the curves 

2x = x'^—xy + a^, x' + y* + 5x = 7y, 

intersect at the origin. Ans. tan <^ = 1.4. 



CHAPTER III 



APPLICATIONS 

1. Tangents and Normals. By the tangent line, or simply 
the tangent, to a curve at any one of its points, P, is meant the 

straight line through P 

X 




whose slope is the same as 
that of the curve at that 
point. 

Let the coordinates of F 
be denoted by (xq, y^. Now, 
the equation of the straight 



Fro. 9 
line through P, whose slope is \, is 

y — ya = K^ — ai))- 

On the other hand, the slope of the curve at any point is Djf. 
If we denote the value of this slope at (a!o, ^o) ^7 {P^\, this 
will be the desired value of A. : 

Hence the equation of the tangent to the curve 

y=f(x) or F{x,y) = 
at the point (aJo, y^ is 

(1) y-yo= (Ps)aip - Xo). 

Since the normal is perpendicular to the tangent, its slope, 
X', is the negative reciprocal of the slope of that line, or 

1 



X'=- 



46 



APPLICATIONS 47 

Hence the equation of the normal to the curve at (x^, y^) is 
(2) 2^-2^o=-Tp-y(aJ-!eo) or x-Xo+iDjy)o- {y -yo)=0. 

Example 1. To find the equation of the tangent to the 

curve 

y-a? 

in the point x = ^,y — \- Here 

Hence the equation of the tangent is 

y — i = ^(x — ^) or 3a! — 4?/ — 1 = 0. 

Example 2. Let the curve be an ellipse : 

t + t = l 
a^ V 

Differentiating the equation as it stands, we get ; 

a^ b^ ah) 

Hence the equation of the tangent is 

This can be transformed as follows : 

a^o2/ - aW = - 6%» + ^'^^, 
V%fB + ahf^y = oV + 6V = a^l^, 



48 CALCULUS 

EXERCISES 

J 1. Find the equation of the tangent of the curve 

y = a? — X 

at the origin ; at the poiat where it crosses the positive axis 
of X. Ans. x + y = 0; 2x — y — 2 = Q. 

2. Find the equation of the tangent and the normal of the 

circle 

a;2 + y2 = 4 

at the point (1, VS) and check your answer. 

3. Show that the equation of the tangent to the hyperbola 

at the point {xq, yg) is 

a" ¥ 

4. Show that the equation of the tangent of the parabola 

y'^ = 2mx 
at the point (xo, yg) is 

VoV = m(» + ai))- 

5. Show that the equation of the tangent of the parabola 

y^ = m? — 2mx 
at the poiat (xg, yg) is 

ygy = m? — m{x +Xg). 

6. Show that the equation of the tangent of the equilateral 

hyperbola 

xy =ia^ 
at the point (xg, yg) is 

y^ + x^ = 2a''. 

7. Find the equation of the tangent to the curve 

a? + y^ = a%x — y) 
at the origin. Ans. x = y. 



APPLICATIONS 



49 



8. Show that the area of the triangle formed by the coordi- 
nate axes and the tangent of the hyperbola 

at any point is constant. 

9. Find the equation of the tangent and the normal of the 

curve . . „ 

a? = a'y' 

in the point distinct from the origin in which it is cut by the 
bisector of the positive coordinate axes. 

10. Show that the portion of the tangent of the curve 

xi + yi = a^ 

at any point, intercepted between the coordinate axes, is 
constant. 

11. The parabola y^ = 2 ax cuts the curve 

a^ — 3 axy + ?/'== 

at the origin and at one other point. Write down the equa- 
tion of the tangent of each curve in the latter point. 

12. Show that the curves of the preceding question intersect 
in the second point at an angle of 32° 12'. 

2. Maxima and Minima. Problem. From a piece of tin- 
3 ft. square a box is to be made by cutting out equal squares 
from the four corners and 
bending up the sides. Detei;-. 
mine the dimensions of the 
box of this description which 
will hold the most. 



^ 



^' 



}- 



3— 2a; 



3— 2a; 



Fio. 10 



Solution. Let x be the 
length of the ■ side of the 
square removed; then the 
dimensions of the box are as indicated in the diagrams, 
noting the cubical content of the box by u, we have : 



De- 



50 



CALCULUS 



1) 



u = x(3-.2xy, 



2) 



u = 9x — 12a;'' + 4*'. 



The problem is, then, to find the value of x which makes u 
as large as possible, x being restricted from the nature of the 
case to being positive and less than f : 

3) 0<a!<f. , 

The problem can be treated graphically by plotting the 
curve 1). We wish to find the highest point on this curve. 

It appears to be the point 
for which x = ^, u = 2, 
since other values of x 
which have been tried lead 
to smaller values of u. 

The foregoing method 
has the advantage that it 
is direct, for it assumes no 
knowledge of mathematics 
beyond curve plotting. It 
has the disadvantage that 
curve plotting, even in the 
simplest cases, is labori- 
ous ; and, furthermore, we 
have not really proved 
that x = ^ is the best 
value. We have merely failed to find a better one. 

The Calculus supplies a means of meeting both the difficul- 
ties mentioned, and yielding a solution with the greatest ease. 
The problem is to find the highest poiat on the curve. At 
this point, the tangent of the curve is evidently parallel to 
the axis of x. Consequently, the slope of the tangent, i.e. 
tan T = D^u, must have the value here : 




Fig. 11 



APPLICATIONS ^ 51 

All we need do, therefore, is to compute D^u, most con- 
veniently from equation 2), and set the result equal to : 

Bji = 9 - 24a! + 12a;2 = 0. 

On solving this quadratic equation for x, we find two roots, 

•'' . — "J* T- 

Only one of these, however, lies within the range 3) of possible 
values for x, namely, the value x = ^, and hence this is the 
required value. 

EXERCISES 

1. Work the foregoing problem for' the case that the tin is 
a rectangle 1 by 2 ft. 

Plot accurately the graph, taking 10 cm. as the unit, and 
determine in this way what appears to be the best value for x, 
correct to one eighth of an inch. 

Solve the problem by the Calculus, and show that the best 
value for x is .21132 ft., or 2.5359 in. 

2. A farmer wishes to fence off a rectangular pasture along 
a straight river, one side of the pasture being formed by the 
river and requiring no fence. He has barbed wire enough to 
build a fence 1000 ft. long. What is the area of the largest 
pasture of the above description which he can fence off ? 

3. Show that, of all rectangles having a given perimeter, 
the square has the largest area. 

4. Show that, of all rectangles having a given area, the 
square has the least perimeter. 

5. Each side of a shelter tent is a rectangle 
6 X 8 ft. How must 1;he tent be pitched so as to 
afford the largest amount of room inside ? The ~^^ ^ 
ends are to be open. 

Ans. The angle along the ridge-pole must be a right 
angle. 




52 CALCULUS 

6. Divide the number 12 into two parts such that the sum 
of their squares may be as small as possible. 

(What is meant is such a division as this : one part might 
be 4, and then the other would be 8. The sum of the squares 
would here be 16 + 64 = 80.) 

7. Divide the number 8 into two such parts that the sum 
of the cube of one part and twice the cube of the other may-be 
as small as possible. 

8. Divide the number 9 into two such parts that the 
product of one part by the square of the other may be as largd^^ 
as possible. 

9. Divide the number 8 into two such parts that the product 
of one part by the cube of the other may be as large as possible. 

10. At noon, one ship, which is steaming east at the rate 
of 2^ mUes an hour, is due south of a second ship steaming 
south at 16 miles an hour, the distance between them being 
82 miles. If both ships hold their courses, show that they 
will be nearest to each other at 2 p.m. 

11. If, in the preceding problem, the second ship lies to 
from noon till one o'clock, and then proceeds on her southerly 
course at 16 miles an hour, when will the ships be nearest to 
each other? 

12. Find the least value of the function 

y = x^ + 6x + 10. Ans. 1. 

cyi3. What is the greatest value of the function 
y = 3x — a^ 
for positive values of a; ? 
-14. -For what value of x does the function 

12Vg 

l + 4a; 

attain its greatest value ? Ans. x = ^ 



APPLICATIONS 



53 



15. At -what point of the interval a < a; < 6, a being posi- 
tive, does the function 



attain its least value ? 



{x— a){h — x) 



Ans. X = Va6. 



1 C*^ 




16. Find the most advantageous length for a lever, by 

means of ■which to raise a 

weight of 490 lb. (see Fig. 13), q 

if the distance of the weight 

from the fulcrum is 1 ft. and the t \qx 

lever weighs 6 lb. to the foot. „ 

° Fig. 13 

i^^ 3. Continuation: Auxiliary Variables. It frequently, — in 
fact, usually, — happens that it is more convenient to formu- 
late a problem if more variables are introduced at the outset 
than are ultimately needed. The following examples will 
serve to illustrate the method. 

Example 1. Let it be required to find the rectangle of 
greatest area which can be inscribed in a given circle. 

It is evident that the area of the 
rectangle will be small when its alti- 
tude is small and also when its base 
is short. Hence the area will be 
largest for some intermediate shape. 
Let u denote the area of the rec- 
tangle. Then 

(1) M = 4 xy. 

But X and y cannot both be chosen 

arbitrarily, for then the rectangle 

will not in general be inseriptible in the given circle. In fact, 

it is clear from the Pythagorean Theorem that x and y must 

satisfy the relation : 

(2) x^ + y^ = A 

We could now eliminate y between equations (1) and (2), 
thus obtaining u in terms of x alone ; and it is, indeed, im- 



Fio. 14 




54 CALCULUS 

portant to think of this elimination as performed, for there is 
only one independent variable ia the problem. The graph of 
u, regarded as a function of x, starts at the origia, rises as x 
increases, but finally comes back to the axis of x again when 
x — a. All this we read off, either from the 
meaning of u and x in the problem or from 
_* equations (1) and (2). 
J. „ It is better, however, in practice not to elimi- 

nate y, but to differentiate equa;tions (1) and 
(2) with respefct ta x as they -Itand, and then set D,,u = 0. 
Thus from (1), " ") 

D^u = 4:(^+xD,y)=0, 

and from (2), 2a; + 2 yD,y = 0. 

From the second of these equations we see that 

y 

Substituting for Dj) this value in the first equation, we get : 
y =0 or y^ = a?. 

y 

Since x and y are both positive numbers, it follows that 

y = x. 
Hence the maximum rectangle is a square. <X 



EXERCISES 

1. Work the same problem for an ellipse, instead of a 
circle. 

2. Work the problem for the case of a variable rectangle 
inscribed in a fixed equilateral triangle. 

Example 2. To find the most economical dimensions for a 
tin dipper, to hold a pint. 




APPLICATIONS 55 

Here, the amount of tin required is to be as small as pos- 
sible, the content of the dipper being given. Let u denote the 
surface, measured in square inches. Then 



a) M = 2irrh + irr^. 

But r and h caimot both be chosen arbitrarily, for 

then the dipper would not in general hold a pint. 

If V denotes the given volume, measured in cubic Fig. 16 

inches, then, since this volume can also be expressed as jrr^, 

we have 

b) irr^A = V. 

Differentiating equation o) with respect to r and setting 
D,ii = 0, we have : 

D,u = ir\2h + 2rDJi + 2 r j = 0, 
or 

c) h + rD,h + r = 0. 
DifEerentiating 6) we get : 

d) T\2rh + rWM=0. 

Now, r cannot = in this problem, and so we may divide this 
last equation through by r, as well as by ^ : 

e) 2h + rD,h, = 0. 

It remains to eliminate D^h between equations c) and e). 
From e), 

r 

Substituting this value of D,A in c), we find : 
/) h—2h + r = 0, or r = h. 

Hence the depth of the dipper must just equal its radius. 

Discussion. Just what have we done here ? The steps we 
have taken are suggested clearly enough by the solution of 



56 CALCULUS 

Example 1. We have chosen one of the two variables, r and 
h, as the independent variable (here, r); differentiated the 
function u, which is to be made a minimum, with respect to r, 
and set D,m = 0. Then we differentiated the second equation 
6), likewise with respect to r, eliminated D,u, and solved. 
But what does it all mean ? What is behind it all.? 

Just this : the quantity m, m the nature of the case, is a 
function of r. For, when to r is given any positive value, a 
dipper can be constructed which will fulfill the requirements. 
Now, if r is very large, we shall have a shallow pan, and evi- 
dently the amount of tin required to make it will be large ; — 
i.e. u will also have a large value. 

But what if r is small ? We shall then have a high cylinder 
of minute cross sections, i.e. a pipe. Is it clear that m, the 
surface, will be large in this case, too? I fear not, for it' is 
purely a relative question as to how high such a pipe must be 
to hold a pint, and I see no way of guessing intelligently. 
By means of equation 6), however, we see that 

and if we substitute this value ma), we get 

Ml / „ F ■ 2F 

Ttr^ r 

From this last formula it is clear that, 
when r is small, u actually is large. 

The graph of u, regarded as a func- 
tion of r, is therefore ia character as 

!^ shown by the accompanying figure. 

It is a continuous curve lying above 
the axis of r, very high when r is small, 
and also very high when r is large. It has, therefore, a 
lowest point, and for this value of r, the area u of the dipper 
will be least. But at this lowest point the slope of the curve, 
i),M, has the value 0. Thus we see, first, that we have a genu- 



O 




APPLICATIONS 57 

ine minimum problem ; — there is actually a dipper of small- 
est area. Secondly, equations c) and d) must hold, and since 
from these equations it follows by elimination that r = ft, there 
is only one such dipper, and its radius is equal to its altitude. 
The problem is, then, completely solved. 

We inquired merely for the shape of the dipper. If the size 
had been asked for, too, it could be found by solving b) and /) 
for r and h, and expressiag Via cubic inches : 

F= ^^ = 28.875, 

fl-r-s = 28.87, ' r = 2.095. 

It can happen in practice that a function attaias its greatest 
or its least value at the end of the interval. In that case, the 
derivative does not have to vanish. Usu- 
ally, the facts are patent, and so no special 
investigation is needed. But it is neces- 
sary to assure oneself that a given problem 
which looks like one of the above does not 
come under this head, and this is done, as 
in the cases discussed in the text, by show- 
ing that near the ends of the interval the values of the func- 
tion are larger, for a minimum problem, than for values well 
withia the interval. 

EXERCISE 

Discuss in a similar manner the best shape for a tomato can 
which is to hold a quart. Here, the tin for the top must also 
be figured in. Show that the height of such a can should be 
equal to the diameter of the base. As to the size of the can, 
its height should be 4.19 in. 

A General Remark. It might seem as if the method used in 
the solution of the above problems were likely to be insecure, 
since the graph of such a function u might, in the very next 
problem, look like the accompanying figure. In such a case, 
there would be several values of x, for each of which D^u = 0, 




58 CALCULUS 

and we should not know which one to take. Curiously enough, 
this case does not arise in practice, — at least, I have never 
come across a physical problem which 
led to this difficulty. In problems 
like the above, there must be at least 
one X for which Djt = ; and when 
we solve a given problem, we actually 
find only one x which fulfills the con- 
Fio. 19 dition. Thus there is no ambiguity. 



EXERCISES 

1. A 300-gallon tank is to be built with a square base and 
vertical sides, and is to be lined with copper. Find the most 
economical proportions. 

Ans. The length and breadth must each be double the 
height. 

T~ 

2. Find the cylinder of greatest volume which 
can be inscribed in a given cone of revolution. ji 

Ans. Its altitude is one-third that of the cone. I 



3. What is the cylinder of greatest convex 



I_ 



surface that can be inscribed in the same cone ? ^" 



Ans. Its altitude is half that of the cone. 



Fio. 20 



4. Of all the cones which can be inscribed in a given sphere, 
find the one whose lateral area is greatest. 

Ans. Its altitude exceeds the radius of the sphere by 33^ % 
of that radius. 

5. Find the volume of the greatest cone of revolution which 
can be inscribed in a given sphere. 

6. If the top and bottom of the tomato can considered in 
the Exercise of the text are cut from sheets of tin so that a 
regular hexagon is used up each time, the waste being a total 
loss, what wiD then be the most economical proportions for 
the can ? 



APPLICATIONS 59 

7. If the strength, of a beam is proportional to its breadth 
and to the square of its depth, find the shape of the strongest 
beam that can be cut from a circular log. 

Ans. The ratio of depth to breadth is V2. 

8. Assuming that the stiffness of a beam is proportional to 
its breadth and to the cube of its depth, find the dimensions 
of the stiffiest beam that can be sawed from a log one foot in 
diameter. 

9. What is the shortest distance from the poiut (10, 0) to ' 
the parabola w2 = 4a;9 

a / ■ 

10. What points of the curve 

are nearest (4, 0) ? 

11. A trough is to be made of a long rectangular-shaped 
piece of copper by bending up the edges so as to give a rec- 
tangular cross-section. How deep should it be made, in order 
that its carrying capacity may be as great as possible ? 

12. Assuming the density of water to be given from 0° to 
30° C. by the formula 

p = po(l + ca + pt^ + yf), 
where po denotes the density at freezing, t the temperature, 
and 

a=5.30xl0-^ /8 = - 6.53 X 10-6, y= 1.4x10-8, 
show that the maximum density occurs at t = 4.08°. 

13. Tangents are drawn to the arc of the ellipse 

E! + ^ = 1 

which lies in the first quadrant. Which one of them cuts off 
from that quadrant the triangle of smallest area ? 

14. Work the same problem for the parabola 

y^ = a^ — iax. 



60 



CALCULUS 



15. Show that, of all circular sectors having the same perim- 
eter, that one has the largest area for which the sum of the 
two straight sides is equal to the curved side. 



A 



y 



o 




o 

"Fia. 21 




4. Increasing and Decreasing Functions. The Calculus 
affords a simple means of determining whether a function is in- 
creasrag or decreasing as the independent variable increases. 

Since the slope of the 
graph is ^iven by D^, 
we see that when D,y 
is positive, y iacreases 
as X iacreases, but 
when Dj/ is negative, 
y decreases as x ia- 
creases. Figure 21 shows the graph ia general when D^ is 
positive. 

In each figure both x and y have been taken as positive. 
But what is said above in the text is equally true when one or 
both of these variables are negative; for the words increase 
and decrease as here used mean algebraic, not numerical, ia- 
crease or decrease. Thus if the temperature is ten degrees below 
zero (i.e. — 10°) and it changes to eight below (— 8°), we say 
the temperature has risen. If 
we measure the time t, in hours 
from noon, then 10 a.m. will 
correspond to t = — 2. Let u 
denote the temperature, meas- 
ured iu degrees. Then a tem- 
perature chart for 24 hours 
from midnight to midnight might look like the accompanying 
figure. At any instant, t = t', for which the slope of the curve, 
D,u, is positive, the temperature is rising, no matter whether 
the thermometer ia above zero or below, and no matter whether 
t is positive or negative ; and similarly, when Z>,m is negative, 
the temperature is falling. 

Again, suppose the amount of business a department store 



t = lZ 




^o. 22 



APPLICATIONS 61 

does in a year, as represented by the net receipts each, day, be 
plotted as a curve (y = receipts, measured in dollars ; x = 
time, measured lq days), the curve being smoothed ia the 
usual way. Then a point of the curve at -which the derivative 
is positive (i.e. D^y > 0) indicates that, at that time, the busi- 
ness of the firm was increasing ; whereas a point at which 
D^ < means that the business was falling off. 

We can state the result in the form of a general theorem, 
.the proof of which is given by inspection of the figure (Fig. 21) 
and the other forms of the figure, brought out in the above 
discussion. 

Theorem : When x increases, then 

(a) if Djf > 0, y increases; 

(b) if ' ' D^y < 0, y decreases. 

Application. As an application consider the condition that 
a curve y =f{x) have its concave side turned upward, as ia 
Fig. 23. The slope of the curve is 
a function of x : 




FlQ. 23 



tan T = ^(w). 

For, when x is given, a point of 
the curve, and hence also the *-* 
slope of the curve at this point, is 
determined. Consider the tangent line at a variable point P. 
If we think of P as tracing out the curve and carrying the 
tangent along with it, the tangent will turn in the counter 
clock-wise sense, the slope thus increasing algebraically as x 
increases, whenever the curve is concave upward. And con- 
versely, if the slope increases as x increases, the tangent will 
turn in the counter clock-wise sense and the curve will be con- 
cave upward. Now by the above theorem, when 

Z>^tanT>0, 

tauT increases as x increases. Hence the curve is concave 
upward, when D, tan t is positive ; and conversely. 



62 



CALCULUS 



The derivative D^tanr is the derivative of the derivative 
of y. This is called the second derivative of y, and is denoted 
as follows : ^ . „ . -r, „ 

(read : " D x second of y ").* 

The test for the curve's being concave downward is obtained 
in a similar manner, and thus we are led to the following 
important theorem. 

Test fob a Cukve's being Concave Upward, etc. The 
curve ,, , 

is CONCAVE UPWARD whsn D^ > ; 

CONCAVE DOWNWARD when D^ < 0. 

A point at which 
the curve changes 
from being concave 
upward and be- 
comes concave 
downward (or vice 
versa) is called a 
poirU of inflection. 
Since D^y changes 
sign at such a point, 

this function will' necessarily, if continuous, vanish there. 

Hence : 

A necessary condition for a point of inflection is that 
BJ'y = 0. 

Example. Consider the curve 

y = 3? — 3x. 

* The derivative of the second derivative, Dx^DJiy), is called the third 
derivative and is written D^^, and so on. 




Fig. 24 



APPLICATIONS 



63 



Its slope at any point is given .by the equation 

The second derivative of y with 
respect to x has the value 

D/2/ = 6a;. 

Thus we see that this curve is 
concave upward for all positive 
values of x, and concave down- ^la. 25 

ward for all negative values. In 
character it is as shown in the accompanying figure. 




EXERCISES 

For what values of x are the following functions increasing ? 
For what values decreasing ? 

1. 2/ = 4 — 2a;2. 

2. y = x^ — 2x + Z. 

Ans. Increasing, when a; > 1 ; decreasing, when x < 1. 

3. 2/ = 5 + 12x — a;2. 

4. y=3fi — 21x + l. 

Ans. Increasihg, when as > 3, and when a; < — 3 ; decreas- 
ing, when — 3 < a; < 3. 

^. y = 5 -\-&x — 3?. 6. y = x — x". 

7. y = ofi — ^x^ + 12x — l. 

In what intervals are the following curves concave upward ; 
in what, downward ? 

8. y = a^ — 3a;2 + 7a; — 5. 

Ans. Concave upward, when a; > 1 ; concave downward, 
when a; < 1. 



64 CALCULUS 

9. 2/ = 15 + 8a; + 3a!= — 05*. 10. y = a^ — Qx^ — x — 1. 

11. y = 3 — 9x + 24:x^—ia^. 12. y = 2a>> — X*. 

13. y = as* — 4.0^ — 6x + 11. 14. y ==-121x + 7x^ — x\ 
15. y = 13 + 23x-24:X'' + 12a^-x^. 

5. Curve Tracing. In the early work of plotting curves 
from their equations the only way we had of finding out what 
the graph of a function looked like was by computiug a large 
number of its points. We are now in possession of powerful 
methods for determining the character of the graph with 
scarcely any computation. I"or, first, we can find the slope of 
the curve at any point; and, secondly, we can determine in 
what intervals the curve is concave upward, in what concave 
downward.* 

Example. Let it be required to plot the curve 

(1) 3y = x^- 3x^ + 1. 

a) Determine first its slope at any point : 

(2) 3n^ = 3x^-6x, Djj = v? — 2x. 

* There are two great applications of the graphical representation of a 
function. One is quantitaMvie, the other, quaUtative. By the first I mean 
the use of the graph as a table, for actual computation. Thus in the use 
of logarithms it is desirable to have a graph of the function y = logio a 
drawn accurately for values of x between 1 and 10 ; for by means of such 
a graph the student can read ofi the logarithms he is using, correct to two 
or three significant figures, and so obtain a check on his numerical vork. 

Therfe is, however, a second large and important class of problems, in 
which the character of a function is the important thing, a minute deter- 
mination of its values being in general irrelevant. 

A case in point is the determination of the number of roots of an alge- 
braic equation, e.g. ss _ ^2 - 4 x -t- 1 = 0, 
Here, we plot the curve y ^^.^ _xi - 4:x + 1 

and inquire where it cuts the axis of x. For this purpose it is altogether 
adequate to know the character of the curve, and for treating this problem 
the methods of the present paragraph yield a powerful instrument. 



APPLICATIONS 



65 



It is always useful to know the points at which the tangent 

to the curve is parallel to the axis of x. These are obtaiued 

hy setting D^y = and solving. Thus we get from (2) the 

equation : 

a;2 — 2a; = 0. 

The roots of this equation are 

x = and x = 2 

Now determine accurately the points having these abscissas, 
plot them, and draw the tangents there : 

y 

We do not yet 
know whether the 
curve lies above its 
tangent in one of 
these points, or be- 
low its tangent ; it 
might even cross 
its tangent, for the 
point might be a 
point of inflection. These questions will all be answered by 
aid of the second derivative. 

6) Compute the second derivative : 

DJh/ = 2a; - 2 = 2(!B - 1). 

We see that it is positive when x is greater than 1 and nega- 
tive when X is less than 1 : 




Fib. 26 



-D.V > 


when 


l<x; 


D^^y < 


when 


X<1. 


DJh, = 


when 


x=l. 



Hence the curve has a point of inflection when x=l. This 
is a most important point on the. curve. We will compute its 



66 CALCULUS 

coordinates accurately, determine the slope of the curve there, 
and draw accurately the tangent there. 

yU=. = -i; -D^U=i — 1. 

This is the last of the important tangents which we need to 
draw. Since the curve is concave upward to the right of the 
liue X = 1, and concave downward to the left of that line, it 
must be in character as indicated. We see, then, that it cuts 
the axis of x between and 1, and again ta the right of the 
poiut a; = 1 ; and it cuts that axis a third time to the left of 
the origin. 

These last two points can be located more accurately by 
computing the function for a few simple values of x. 

hence the curve cuts the axis of x between a; = 2 and a; = 3. 

2/L=-i=-3; 

hence the curve cuts the axis between a; = and a; = — 1. 
Incidentally we have shown that the cubic equation 

a? — 3x^ + 1 = 

has three real roots, and we have located each between two 
successive integers. 

EXERCISES 

Discuss in a similar manner the following curves. In par- 
ticular : 

a) Determine the points at which the tangent is horizontal, 
if such exist, and draw the tangent at each of these points ; 

6) Determine the intervals in which the curve is concave 
upward, and those in which it is concave downward ; 

c) Determine the points' of inflection, if any exist, and draw 
the tangent in each of these points ; 



APPLICATIONS 67 

d) Draw in the curve.* 

In most cases it is desirable to take 2 em, as the imit. 

1. y = x^ + 3x^ — 2. 

2. y = a^—3x + l. 

3. y=iifl +3x + l. 

4. 6y = 2a^-3x^-12x + 6. 

5. 6y = 2d' + 3x^-12x — 4:. 

6. y = a? + x^ + x + l. 

Suggestion. Show that the derivative has no real roots and 
hence, being continuous, never changes sign. 

7. 12y = 4:a?-6x^ + 12x — 9. 

8. y = 2x' — x — a^. 

9. 12y = 4a!8 + 18a!2 + 27a; + 12. 

10. 3/ = l — 4a! + 6c(;2_3ffi3. 

11. y = l + 2x + x^ — 0(?. 

12. 42/ = a!* — 6a!2 + 8. 13. y = a^ — ^ix"^ + 1. 
14. y = x — a^. 15. y = X + x^. 

16. y = a!* + a;2. 17. y=2xl*~xK 

18. y = 3a!5 + 6a!' + 15a! + 2. 

19. 60y = 23^ + 16a^ + GOaj^ — 30. 

6. Relative Maxima and Minima. Points of Inflection. A 

function 

(1) • 2/ =/(«') 

* Since a curve separates very slowly from its tangent near a point of 
inflection, the material graph of the curve must necessarily coincide with 
the material graph of the tangent for some little distance. 



68 CALCULUS 

is said to have a maximum at a point x = Xoii its value at Xq is 
larger than at any other point in the neighborhood of a%. But 
such a maximum need not represent the largest value of the 
function in the complete interval a ^ a; ^ 6, as is shown by- 
Fig. 27, and for this reason 
it is called a relative maxi- 
mum, in distinction from 
a maximum maximorum, 
X or an absolute maximum. 
a;=a x„ x^ x=b A. similar definition 

^°- ^ holds for a minimum, the 

word " larger " merely being replaced by " smaller." 

It is obvious that a characteristic feature of a maximum is 
that the tangent there is parallel to the axis of x, the curve 
being concave downward. Similarly for a minimum, the curve 
here beiag concave upward. Hence the foUowiag 

Test fok a Maximum ob a Minimum. If 

(a) [Z)^]_=0, [i).^]_„<0, 

the function has a maximum for x = Xo; if 

(6) iDjy-]^=0, [i).^]_„>0, 

it has a minimum. 

The condition is sufficient, but not necessary ; cf . § 7. 

Example. Let y = !ic^ — 3x'' + l. 

Here D^ = Gaj^- 6a; = 6a;(a!2- l)(a!= + 1), 

and hence D^y = for a; = — 1, 0, 1. 

Thus the necessary condition for a maximum or a minimum, 
Dj/ = 0, is satisfied at each of the points a; = — 1, 0, 1. 

To complete the determination, if possible, compute the 
second derivative, j)^^^s0<^-6, 

and determine its sign at each of these points : 




Fio. 28 



APPLICATIONS 69 

[i)j^]j=_i = 24 > 0, .-. x = —l gives a minimum; 
[D,^2/],_o =— 6<0, .•.x= gives a maximum ; 
[PJh/],,^ = 24 > 0, .-. x= 1 gives a minimum. 

Points of Inflection. A point of inflection is characterized 
geometrically by the phenomenon that, as a point P describes 
the curve, the tangent at P 
ceases rotating in the one di- y 
rection and, tumiag back, be- 
gins to rotate in the opposite 
direction. Hence the slope 
of the curve, tan t, has either 
a maximum or a minimum at — 
a point of inflection. 

Conversely, if tan r has a 
maximum or a minimum, the curve will have a point of inflec- 
tion. For, suppose tan t is at a maximum when x = Xg. Then 
as X, starting with the value Xq, increases, tan t, i.e. the slope 
of the curve, decreases algebraically, and so the curve is con- 
cave downward to the right of a^. On the other hand, as x 
decreases, tan t also decreases, and so the curve is concave up- 
ward to the left of X(,. 

Now, we have just obtained a theorem ■W'hich insures us a 
maximum or a minimum in the case of any function which 
satisfies the conditions of the theorem. If, then, we choose 
as that function, tanr, the theorem tells us that tanr wUl 
surely be at a maximum or a minimum if 

D,tanr = 0, DJ^ tsmr^O. 

Hence, remembering that 

tan T = D^y, 
we obtain the following 

Test fok a Point of Inflection. If 
the curve has a point of inflection at x— Xf,. 



70 



CALCULUS 



This test, like the foregoLag for a maxionim or a minimum, 
is sufficient, but not necessary ; cf. § 7. 
Example. Let 

. 272/ = ffi* + 2a!S-12a;2 + 14a; — 1. 
Then 27D,y = 4:afi + 6x' — 2ix + U, 

27 DJh/ = 12a:'' + 12 a! - 24 = 12(a! - l)(a; + 2), 
2rZ),52/ = 12(2a; + l). 

Setting D/y = 0, we get the points a; = 1 and x = — 2. And 

S1I1C6 

27[A'2/]^= 36 =ifc 0, 27[X>,'2^]_., 36 =jfc 0, 

we see that both of these points are points of inflection. 

The s,lope of the curve in these points is given by the 
equations : 



27[i?^]^i=0, 



27[Z)^]_3=54. 



Hence the curve is parallel to the axis of x at the first of these 
points ; at the second its slope is 2. 



EXERCISES 

Test the following curves for maxima, minima, and points 
of inflectioil, and * determine the slope of the curve in each 
point of inflection. 

1. y =: 4:0^— 15x^+12 x+1.- 3. 6y = afi — Ssc'+Sa^—l. 

Z. y = !i^ + a^ + iifi. 4. y=(x-iy{x+2y. . 




^- y=-, 



2 + 3a? 

6. y=(l — x^^ 

7. Deduce a test 
for "distinguishing be- 
tween two such points of inflection as those indicated in 
Fig. 29. 



APPLICATIONS 71 

7. Necessary and Sufficient Conditions. In order to under- 
stand the nature of the tests obtained in the foregoing paragraph 
it is essential that the student have clearly in miad the mean- 
ing of a necessary condition and of a sufficient condition. Let 
us illustrate these ideas by means of some simple examples. 

a) A necessary condition that a quadrilateral be a square is 
that its angles be right angles. But the condition is obviously 
not sufficient ; all rectangles also satisfy it. 

6) A sufficient condition that a quadrilateral be a square is 
that its angles be right angles and each side be 4 in. long. 
But the condition is obviously not necessary | the sides might 
be 6 iu. long. 

c) A necessary and sufficient condition that a quadrilateral 
be a square is that its angles be right angles and its sides be 
mutually equal. 

As a further illustration consider the following. It is a 
well-known fact about whole numbers that if the sum of the 
digits of a whole number is divisible by 3, the number is divis- 
ible by 3 ; and conversely. Also, if the sum of the digits of a 
whole number is divisible by 9, the number is divisible by 9 ; 
and conversely. Hence we can say : 

i) A necessary condition that a whole number be divisible 
by 9 is that the sum of its digits be divisible by 3. But the 
condition is not sufficient. 

ii) A sufficient condition that a whole number be divisible 
by 3 is that the sum of its digits be divisible by 9. But the 
condition is not necessary. 

iii) A necessary and sufficient condition that a whole num- 
ber be divisible by 3 (or 9) is that the sum of its digits be 

divisible by 3 (or 9). 

t 

Turning now to the considerations of § 6, we see that a 
necessary condition for a miaimum is that 



72 CALCULUS 

at the point in question, a; = a^. But this condition is not 
sufficient. When it is fulfilled, the function may have a 
maximum, or it may have a point of inflection with horizontal 
tangent. 

On the other hand, the condition 

is sufficient for a minimum. But it is not necessary. Thus 
the function 

(1) y=x^ 

ohviously has a minimum when x = 0. The necessary condi- 
tion, Dj/ = 0, is of course fulfilled : 

But here Z>,=2/ = 12 a;^, and [D^^']^ 

is not positive ; it is 0. 

Again, as was shown in § 4, a necessary condition for a point 

of inflection is that ^ „ 

DJhf = 

at that point. But this condition is not sufficient. Thus in 
the case of the curve (1) this condition is fulfilled at the 
origin. But the origin is not a point of inflection. 

Remark. It may seem to the student that such tests are 
unsatisfactory since they do not apply to all cases and thus 
appear to be incomplete. But their very strength lies in the 
fact that they do not tell the truth in too much detail. They 
single out the big thing in the cases which arise in practice 
and yield criteria which can be applied with ease to the great 
majority of these cases. 

8. Velocity; Bates. By the average velocity with which a 
point moves for a given length of time t is meant the distance 
8 traversed divided by the time : 

average velocity = -■ 



APPLICATIONS 73 

Thus a railroad train which covers the distance between two 
stations 16 miles apart in half an hour has an average speed 
of 15/-|^= 30 mUes an hour. 

When, however, the point in question is moving sometimes 
fast and sometimes slowly, we can describe its speed approxi- 
mately at any given instant by considering a short interval 
of time immediately succeeding the instant t^ in question, and 
taking the average velocity for this short interval. 

For example, a stone dropped from rest falls according to 
the law : . „ „ 

To-find how fast it is going after the lapse of to seconds. Here 

(1) So = 16«o'. 

A little later, at the end of t' seconds from the beginning of 
the fall, 

(2) s' = 16«'2 

and the average velocity for the interval of t' — to seconds is 

(3) ^j^^ ft. per second. 

t — to 

Let us consider this average velocity, in particular, after the 

lapse of 1 second : ^ . „ 

to = 1, So •= 16. 

Let the interval of time, t' — to, be -^ sec. Then 

s' = 16 X 1.1== = 19.36, 

s;^-So _ 3^ ^ 336 ft. a second. 
t' - to .1 

Thus the average velocity for one-tenth of a second immedi- 
ately succeeding the end of the first second of fall is 33.6 ft. a 
second. 

Next, let the interval of time be j^ sec. Then a similar 
computation gives, to three significant figures : 

^^^^ = 32.2 ft. a second. 
t' - to 



74 CALCULUS 

And when the interval is taken as x^^nr ^^°-' *^® average 
velocity is 32.0 ft. a second. 

These numerical results iadicate that we can get at the 
speed of the stone at any desired instant to any desired degree 
of accuracy by direct computation ; we need only to reckon 
out the average velocity for a sufB.eiently short interval of 
time succeeding the iastant in question. 

We can proceed in a similar manner when a point moves 
according to any given law. Can we not, however, by the aid 
of the Calculus avoid the labor of the computations and at the 
same time make precise exactly what is meant by the velocity 
of the point at a given instant? If we regard the interval 
of time- 1' — io as an increment of the variable t and write 
«' — <o = Ai, then s' — So = As will represent the corresponding 
increment in the function, and thus we have : 

average velocity = — 

Now allow At to approach as its limit. Then the average 
velocity will in general approach a limit, and this limit we take 
as the definition of the velocity, v, at the instant ^ : 

lim (average velocity from t=tgto t = f) 

= actual velocity * at instant t = to, 

or v = lim — = D,s. 

At^ At 

Hence it appears that the velocity of a point is the time- 
derivative of the space it has traveled. In the ease of a 
freely falling body this velocity is 

v = D,s = 32 1. 

In the foregoing definition, s has been taken as the distance 
actually traversed by the moving point, P. More generally, 
let s denote the length of the arc of the curve on which P is 
moving, s being measured from an arbitrarily chosen fixed 

* Sometimes called the instantarieous velocity. 



APPLICATIONS 75 

point of tliat curve. Either direction along the curve may be 
chosen as the positive sense for s. Thus, in the case of a 
freely falling body, s might be taken as the distance of the 
body above the ground. If h denotes the initial distance, then 

where s' denotes the distance actually traversed by P a' 
at any given instant. Hence , 

D,s + D/ = 0, ^ s 



or D,s = — D,s'. Fig. 30 

Here D^ gives numerically the value of the velocity, but D^s is 
a negative quantity. 

We will, accordingly, extend the conception of velocity, 
defining the velocity v of the point as D,s : 

V = DfS. 

Thus the numerical value of v or D,s will always give the 
speed, or the value of the velocity in the earlier sense. In 
case s increases with the time, DfS is positive and represents 
the speed. If, however, s decreases with the time, D,s is nega- 
tive, and the velocity, v, is therefore here negative, the speed 
now being given by — « or — X),s. In all eases, 

Speed=|u| = \D,s\. 

Example. Let a body be projected upward with an initial 
velocity of 96 ft. a second. Assuming from Physics the law 

that s = 96f-16«2, 

find its velocity a) at the end of 2 sec. 

6) at the end of 5 sec. 

Solution. By definition, the velocity at any instant is 

V = D,s = 96 -32t. 
Hence 

a) v\^,=^. ^V 

6) «l^ = -64. 



76 CALCULUS 

The meaning of these results is that, at each of the two 
instants, the speed is the same, namely, 64 ft. a second (and 
the height above the ground is also seen to be the same, 
s = 128 ft.). But when t = 2, DiS is positive ; hence s is in- 
creasing with the time and the body is rising. When t = 5, 
D,s is negative ; hence s is decreasing with the time, and the 
body is descending. 

Rates. Consider any length or distance, r, which is chang- 
ing with the time, and so is a function of the time. Let j-q 
denote the value of »• at a given Instant, t = to, and let r' be the 
value of r at a later instant, t — t'. Then the increase in r 
will be r' — ro = Ar and that in t will be t' — ta = ^t. Thus in 
the interval of time of At seconds succeeding the instant t = Jj, 

average rate of increase of r = — 

At 

Now let At approach as its limit. Then the average rate of 
increase will in general approach a limit, and this limit we take 
as the definition of the rate of increase of r at the instant t^ : 

lim (average rate of increase from ^ = {^ to f = t') 

= actual rate of increase at instant f = <o 

= lim — = Dir. 

A«=o At 

In other words, the rate at which r is increasing at any in- 
stant is defined as the time-derivative of r. 

If r is decreasing, D,r wUl be a negative quantity ; and con- 
versely, if D,r is negative, then r is decreasing. In either ease, 
the numerical value of D^r gives the rate of change of r ; just 
as, in the case of velocities, the numerical value of D,s gives 
the speed. 

More generally, instead of r, we may have any physical 
quantity, u, as an area or a volume or the current in an electric 
circuit or the number of calories in a given body. 



APPLICATIONS 



77 



In all these cases, the rate at which u is increasing ia defined 
as the time-derivative of u, i.e. as D,u ; and the rate of change 
of M is I D,u |. 

Example. At noon, one ship is steaming east at the rate of 
18 miles an hour, and a second ship, 40 miles north of the first, 
is steaming south at the rate of 20 miles an hour. At what 
rate are they separating from each other at one o'clock ? 



Solution. 


The relation between r and t is 






here given 


by the Pythagorean Theorem : 






or 


r' = (A0-20ty+{lSty, 


1 


\ 


(1) 


r2 = 1600 - 1600* + 724^2. 


i 


\ 


Hence 






18i\ 






(2) 


r = V1600 - IGOOt + 724<2. 


Fig. 31 



We wish to find D,r. This can be done by difEerentiatiag 
equation (2) ; but that would be poor technique, since it is 
simpler to diEEerentiate equation (1) through with respect to t: 



(3) 



2?-i),r = -1600 + 1448<, 

- 800 + 724« 
D,r = 



Equation (3) gives the rate at which r is increasing at any 
instant t ; i.e. t hours past noon, or at t o'clock. 
Setting now, in particular, * = 1, we obtain : 



A»-|.=i=- 



76 



= - 2.825. 



The meaning of this result is twofold. First, since D,r is 
negative when f = 1, the ships are not receding from each 
other, but are coming nearer together. Secondly, the rate of 
change of the distance between them is, at one o'clock, 2.825 
miles an hour. 



78 CALCULUS 

Let the student determine how long they will continue to 
approach each other, and what the shortest distance between 
them will he. 

Remark. It is important for the student to reflect on the 
method of solution of this problem, since it is typical. We 
were asked to find the rate of recession at just one instant, t=l. 
We began by determining the rate of recession generally, i.e. 
for an arbitrary instant, t=t. Having solved the general 
problem^ we then, as the last step -in the process, brought into 
play the specific value of f which alone we cared for, namely, 
t = l. 

The student will meet this method again and again, — in 
integration, in mechanics, in series, etc. We can formulate 
the foregoing remark suggestively as follows : By means of 
the Calculus we can often determine a particular physical 
quantity, like a velocity, an area, or the time it takes a body, 
acted on by known forces, to reach a certain position. The 
method consists in first determining a function, whereby the 
general problem is solved for the variable case ; and then, as 
the last step in the process, the special numerical values with 
which alone the proposed question is concerned, are brought 
into play. 

EXERCISES 

1. The height of a stone thrown vertically upward is given 
bytheformula: s=m-16t^. 

When it has been rising for one second, find (a) its average 
velocity for the next ^ sec. ; (6) for the next ^^ sec. ; (c) its 
actual velocity at the end of the first second ; (d) how high it 
will rise. 

Ans. (a) 14.4 ft. a second ; (6) 15.84 ft. a second ; (c) 16 ft. 
a second ; (d) 36 ft. 

2. One ship is 80 miles due south of another ship at noon, 
and is sailing north at the rate of 10 miles an hour. The 



APPLICATIONS 79 

second ship sails west at the rate of 12 miles an hour. Will 
the ships be approaching each other or receding from ^each 
other at 2 o'clock ? What -will be the rate at which the dis- 
tance between them is changing at that time ? How long will 
they continue to approach each other ? 

3. If two ships start abreast half a mile apart and sail due 
north at the rates of 9 miles an hour and 12 miles an hour, 
how far apart will they be at the end of half an hour ? How 
fast will they be receding at that time ? 

4. Two ships are steaming east, one at the rate of 18 miles 
an hour, the other at the rate of 24 miles an hour. At noon, 
one is 50 miles south of the other. How fast are they sepa- 
rating at 7 P.M. ? 

5. A ladder 20 ft. long rests against a house. A man 
takes hold of the lower end of the ladder and walks off with 
it at the uniform rate of 2 ft. a second. How fast is the upper 
end of the ladder coming down the wall when the man is 4 ft. 
from the house ? 

6. A kite is 150 ft. high and there are 250 ft. of cord out. 
If the kite moves horizontally at the rate of 4 m. an hour 
directly away from the person who is flying it, how fast is the 
cord being paid out ? Ans. 3^ m. an hour. 

7. A stone is dropped into a placid pond and sends out 
a series of concentric circular ripples. If the radius of the 
outer ripple increases steadily at the rate of 6 ft. a second, 
how rapidly is the area of the water disturbed increasing at 
the end of 2 sec. ? Ans. 452 sq. ft. a second. 

8. A spherical raindrop is gathering moisture at such a 
rate that the radius is steadily increasing at the rate of 1 mm. 
a minute. How fast is the volume of the drop increasing 
when the diameter is 2 mm. ? 

9. A man is walking over a bridge at the rate of 4 miles an 
hour, and a boat passes under the bridge immediately below 
him rowing 8 miles an hour. The bridge is 20 ft. above the 



80 CALCULUS 

boat. How fast are the boat and the man separating 3 min- 
utes later ? 

Suggestion. The student should make a space model for 
this problem by means, for example, of the edge of a table, a 

crack La the floor, and a string ; or 
by two edges of the room which do 
not intersect, and a string. He 
should then make a drawing of his 
model such as is here indicated. 

10. A locomotive running 30 
miles an hour over a high bridge 
'"■ dislodges a stone lying, near the 

track. The stone begins to fall just as the locomotive passes 
the point where it lay. How fast are the stone and the loco- 
motive separating 2 sec. later ? * 

11. Solve the same problem if the stone drops from a point 
40 ft. from the track and at the same level, when the locomo- 
tive passes. 

12. A lamp-post is distant 10 ft. from a street crossing and 
60 ft. from the houses on the opposite side of the street. A 
man crosses the street, walking on the crossing at the rate of 
4 miles an hour. How fast is his shadow moving along the 
walls of the houses when he is halfway over ? 

* BScHEB, Plane Analytic Geometry, p. 230. 




CHAPTER IV 
INFINITESIMALS AND DIFFERENTIALS 

1. Infinitesimals. An infinitesimal is a variable which it is 
desirable to consider only for values numerically small and 
which, when the formulation of the problem in hand has pro- 
gressed to a certain stage, is allowed to approach as its limit. 

Thus in the problem of differentiation, or finding the limit 

(1) lim^=Z>^, 

Aa! and Ay are infinitesimals ; for we allow Aa; to approach 
as its limit, and then Ay also approaches 0.. 

Again, if we denote the value of the difference Ay/ Ax — Dj/ 
by £, so that 

(2) f-2>^ = e, 

Ao; 

then e is an infinitesimal. For, when Aa; approaches 0, the 
left-hand side of equation (2) approaches 0, and so e is a vari- 
able which approaches as its limit, i.e. an infinitesimal. 

Principal Infinitesimal, When we are dealing with a num- 
ber of infinitesimals, a, p, y, etc., it is usually possible to 
choose any one of them as the independent variable, the others 
then becomiag functions of it, or dependent variables. That 
infinitesimal which is chosen as the independent variable is 
called the principal infinitesimal. 

Thus, if the infinitesimals are a and jS, and if 

81 



82 CALCULUS 

it is natural to choose « as the principal infinitesimal. But 
it is perfectly possible to take j8 as the principal infinitesimal. 
« then becomes the dependent variable, and is expressed ia 
terms of p by solving equation (3) for a. : 

(4) « = — ^ 

^' 2-3/3 

Order of Infinitesimals. We are going to separate infinitesi- 
mals into classes, according to the relative speed with vrhich 
they approach 0. Suppose we let a. set the pace, taking on 
the values .5, .1, .01, .001, etc. Consider, for example, a'. 
Then «' takes on the respective values .25, .01, .0001, etc., and 
hence runs far ahead of a : 



« 


.5 


.1 


.01 


.001 ■•. 


oC' 


.25 


.01 


.0001 


.000001 - 



Purthermorej the closer the two get to 0, the relatively nearer 
0? is to 0. Thus, when a = .5, a^ is twice as close ; but when 
a = .01, «' is one hundred times as close ; and so on. 

Again, consider the infinitesimal \a. It is always twice as 
close to as a is. Similarly, 10 « is always one-tenth as close 
as a,. 

From these examples we see that there is a decided difference 
between the relative behavior of a and Tea on the one hand, 
and that of a and «2 on the other. For, lea. is keeping pace 
relatively with a, whereas a^ runs indefinitely ahead of «, rela- 
tively. Consequently, we should put lea into the same class 
with «, whereas a? forms the starting point for a new class. 
To this latter class would belong such infinitesimals as ^«' or 
4a' — a' ; and the former class would include, for example, 
2« + 3a- and -fj-j^a — lOOOa*. Let the student make out a 
table like the above for each of these examples. 

What is the common property of all infinitesimals of the 
same class ? Is it not, that, for two infinitesimals, the relative 
speed with which they approach is nearly, or quite, a fixed 
number not zero ? It is this idea which lies at the bottom of 



INFINITESIMALS AND DIFFERENTIALS 83 

the conception of the order of an infinitesimal, and it is for- 
mulated in a precise definition as follows : 

Definition. Two infinitesimals, p and y, are said to be of 
the same order if their ratio approaches a limit not ; 

lim^=ir:^'0. 
V 

Thus ;8=2« + a? and y = 3a — a' 

are of the same order. For, 

j8^ 2ct + «' ^ 2+ tt 
y 3 a — a? 3 — o? 

and hence, when « approaches 0, 

,. fl ,. 2-l-a 2 . 
limS=lim-^ = -=^0. 

Similarly, 12 o? + 5a^ and Ba^ — 7 a' are infinitesimals of the 
same order. 
An infinitesimal /3 is said to be of higher order than y if 

limi2 = 0, 
7 

Thus if j3=9a* and y = 2a + 5a*, 

j8 is of higher order than a. For, 

^^ 9a' _ 9a 
y~2a+5a* 2 + 5a" 

and hence, when a approaches 0, 

,. fl T 9a 

y 2 + 5 «' 

Finally, p is said to be of lower order than y if 
(5) lim"=oD, 

(read: "j8/y becomes infinite" ; not ";8/y equals infinity."*). 

* The student should now turn back to Chapter 11, § 6, and read again 
carefully what is said there ahout infinity. In particular, he should im- 



84 CALCULUS 

Thus if /8=Va and y=6a + a», 

/8 is of lower order than y. For 

;3_ V^ ^ 1 

y 6«+a' Va(6 + a') 

When a approaches 0, it is evident that the last fraction in- 
creases without limit, or „ 

lim^ = oo. 

y 

First Order, Second Oi-der, etc. An infinitesimal /i is said to 
be of the ^rs< order if it is of the same order as the principal 
infinitesimal, a ; i.e. if „ 

a 

If /8 is of the same order as a\ i.e. if 

lim4 = ^¥=0, 

then ^ is said to be of the second order. And, generally, if ff 
is of the same order as a", i.e. if 

limJ?.= ^:^0, 
a" 

then p is said to be of the n4h order. 
Thus if 

/8 = 2a or p = ^— or Q = a + (t\ 
2 — a 
then j8 is of the first order. 

But if 

^ = 2«2+a' or I3 = -J^ or B=a\ 

3 + a 
then fi is of the second order. 

press on his mind the fact that infinity is not a limit and that in the 
notation used in (5) the = sign does not mean that one number is equal 
to another number. The formula is not an equation in the sense in which 
2x = 3 or o' — 62 = (a — 6) (a + 6) is an equation. The formula means 
no more and no less than that the variable /S/7 increases in value without 
limit. 



INFINITESIMALS AND DIFFERENTIALS 85 



If^=Va, then /3 _ . 



J 

and Iim-^=lgfcO. 

a* 
Hence /8 is of the order ^. 

It is easily seen that if two infinitesimals /S and y are, under 
the present definition, each of order n, then they also satisfy 
the earlier definition of being of the same order. For, let 

hm.-^ = K^O and lim-^ = i:^0. 
«■" a" 

Then, if we denote the differences /S/et" — K and y/a" — L 
respectively by e and rj, so that 

(6) K-K=e and ^-i = «, 

«" a" 

these variables, e and r), will be infinitesimals. For, the left- 
hand side of each of the equations (6) approaches 0. 
From equations (6) it follows that 

^ = K+ £ and ^ = Z, + ». 

«" a" 

On dividing one of these equations by the other we have : 

y L + rj 
We are now ready to allow a to approach as its limit. Then 

lim^=lim^^ 
y L + Ti 

By Theorem III of Chapter 2, § 5 this last limit has the value 

lini^±J = i-™L(:K+i) = :^. 
L + ri lim (L + 7i) L 

Hence, finally j^^ ^^^:^o, q. e. d. 

y L 



86 CALCULUS 

EXERCISES 

1. Snow tnat 

^ = 5a— ll«2 + a3 and y = 7a + a< 
are infinitesimals of the same order. 

2. Show that 

/8 = 2«-3a2 and y = 2a + a* 

are infinitesimals of the same order, but that their difference, 
^ — y, is of higher order than yS (or y). 

3. Show that B = is an infinitesimal of the second 

^ a' -2 

order, referred to « as_principal infinitesimal. 



4. Show that /8 = Va^ + 2 a* is of the first order, referred 
to a. ■ 

5. Show that ^ = V2« + 13a' is of lower order than a. 

6. Show that the order of fi in question 5 is n = ^. 

Determine the order of each of the following infinitesimals, 
referred to a as the principal infinitesimal : 

7. |a + 18a3. 11. -s/i^ - a. 

8. -a+V2«' + «*. 12. ^- aW + „i8. 

la!' 



*■ 13 -a ' ^^- v2a2-«'. 



10. 



\8-7« ■ >/ «2 



+ 4tt* 



+ 2 



15. If and y are infinitesimals of orders n and m respec- 
tively, show that their product, ^y, is an infinitesimal of order 
M + m. 

16. If j8 and y are infinitesimals of the same order, show 
that their sum is, in general, an infinitesimal of the same 
order. 

Are there exceptions ? Illustrate by examples. 



INFINITESIMALS AND DIFFERENTIALS 87 

2. Continuation ; Fundamental Theorem. Principal Part of 
an Infinitesimal. Let j8 be an iniiiiiteaimal of order n, and 
let a be the principal infinitesimal. Then 

lim-^ = ir^O. 
cC 

Moreover, as pointed out in the last paragraph, 

(1) 4=^+^' 

a" 

where e is infinitesimal. Prom (1) it follows that 
(2) p = Za" + ea". 

This last equation gives a most important analysis {i.e. break- 
ing up) of /3 into two parts, each of which is simple for its 
own peculiar reason. 

i) Ka'^ is the simplest infinitesimal of the nth. order imagi- 
nable, — a monomial in the independent variable, the function 

y = Kx\ 

ii) cot'' is an infinitesimal of higher order than the nth. 
The first part, ^a", is called the principal part of ^S. 
By far the most important case in practice is that of infini- 
tesimals /8 of the ^rst order, n = 1. Here 

«" a 
and /8 = Ka -\- eot. 

Hence we see that the principal part of an infinitesimal of the 
first order is proportional to the principal infinitesimal. 

Example 1. Let yS = 2a — a^. 

Then p is obviously of the first order, or m = 1, and here 

a" a 



88 CALCULUS 

Clearly, theu, K=2, € = — a, 

and the principal part of p is 2a. 



Example 2. 


Let 


;8 = 


2«- 
7-4a 




Here, obviously, n = 


= 2, and 








- 


lim-^ = 
a^ 


7-4a 


.2 

■7' 


Hence K=-- 

7 


By 


definition 

e = 


a" 




In the present 


case, 


then. 








£ = 


- 2 , 


2 8a 






7-4a 


7 7(7-4 


^) 



EXERCISE 

Determine the principal parts of a goodly number of the 
infinitesimals occurring in the Exercises at the end of § 1. 

Equivalent Infinitesimals. Two infinitesimals, as ^ and y, 
shall be said to be equivalent if the limit of their ratio is unity : 

lim^=l. 

y 

For example, the following pairs of infinitesimals are equiv- 
alent : 

i) 2 a 4- a2 and 2 a + a^ ; 

ii) \a?- — a? and \a^ + a?; 

iii) V2 a + 5 a2 and V2a— 7«^. 

An infinitesimal and its principal part are always equivalent 
infinitesimals. For, if Ea" is the principal part of /8, then 



INFINITESIMALS AND DIFFERENTIALS 89 

■where ■q is of higher order than Ka". Hence 

-^ 1+-^, lim-^=l + limJ-. 

Ka" Ka"' Ka" Ka" 

But lim rj/Ka" — 0, and the statement is established. 

Two infinitesimals which have the same principal parts are 
equivalent, and conversely. 

Equivalent infinitesimals are of the same order ; but the 
converse is not true. 

The difference between two equivalent infinitesimals, /8 and 
y, namely, /3 — y, is of higher order than j8 or y. For 

y y 

hence lim ^-^^ = lim f *? - l'^ 



y 



-(?-) 



= Aim ^V 1 = 0, q.e.d. 

Conversely, if (3 and y are two infinitesimals whose differ- 
ence, /8 — y, is of higher order than /3 or y, then (3 and y are 
equivalent. 

For, since i2^l2 = ^_l, 

y 7 

it follows that j.^/^ _ ^\ li^^sU^. 

The right-hand side of this equation is by hypothesis, and 
the left-hand side is equal to 

Aim^Vl. 

^ yJ 

Hence lim"=l, q.e.d. 

y 

We come now to a theorem of prime importance in the 
Infinitesimal Calculus. 



90 CALCULUS 

Fundamental Theorem. The limit of the ratio of two infini- 
tesimals, 

7 

is unchanged if the numerator infinitesimul j8 be replaaed by any 
equivalent infinitesimal /3' and the denominator infinitesimal y be 
replaced by any equivalent infinitesimal y'. 
In other words : 

lim " = lim c. 

y y 

provided li^l^-L „^ li^^^^l. 

The proof is immediate. It is obvious that 

Hence by Theorem II, Chapter II, § 5 we have 

But the first and third limits on the right-hand side are each 
equal to 1 by hypothesis. Hence 

lim " = lim ^, q. e. d. 

y y 

The theorem can be stated in the following equivalent 
form: 

The limit of the ratio of two infinitesimals is the same as the 
limit of the ratio of their principal parts. 

The student must not generalize from this theorem and 
infer that an infinitesimal can always and for all purposes be 
replaced by an equivalent infinitesimal. Thus if 

P — 2a + a^ and y = 2a — a\ 

their difference, /8 — y = «' + «*, 



lim-'" 



INFINITESIMALS AND DIFFERENTIALS 91 

is an infinitesimal of the second order. On the other hand, 

is equivalent to y. But it is not true that the difference of p 
/3-y' = aS 



and y', namely, , _ 



is an infinitesimal of the second order. It is obviously of 
order 3. Thus replacing y by an equivalent infinitesimal has 
here changed the order of the difference j8 — y. 



^^t yu,c 



-^^/. 




,v^ 3. Differentials. Let y=f(x) 
be a function of x, and let Dj/ be its derivative : 

lim^ = i?^. ^^/^i 

A«=a) Ax ^ : 

Let the difference Ay /Ax — D^y be denoted by e. Then 

, Ax 

and 

(1) Ay = DjyAx + eAx. 

Since x is the independent variable, Ax can be taken as the 
principal infinitesimal. D^y does not vary with Ax; it is a 
constant, for we are considering its value at a fixed point 
X = Xa. Since, moreover, D^y is not in general zero, equation 

(1) represents Ay as the sum of its principal part, DjjAx, and 
an infinitesimal of higher order, €Ax. 

Definition of a Differential. The expression D^yAx is called 
the differential of the function, and is denoted by dy : 

(2) dy = D,yAx, or df(x)=DJ(x)Ax. 

(read : " differential y " or " differential f{x) " or " dy," etc.). 

Thus if y = a?, 

dy — 2xAx, or dx^ = 2xAx. 



92 



CALCULUS 



Since the definition (2) holds for every function y =f(x), it 
can be applied to the particular function 

Hence •^('")=^- 

(3) dx = D^x Ak = Ak. 

But it is not ia general true that i^y and dy are equal, since t 
is in general different from 0. Thus we see that the differen- 
tial of the independent variable is equal to the increment of that 
variable ; but the differential of the dependent variable is not in 
general equal to the increment of thai variable. 

By means of (3) equation (2) can now be written in the form 

(4) dy = D^dx. 



Hence 

(5) 



dx 



Geometrically, the increment A?/ of the function is repre- 
sented by the line MP', Fig. 33 ; and the differential, dy, is 

equal to MQ, for from (5) " 

tan T = — 
dx 

or dy = dx tan t. 

In other words. Ay repre- 
sents the distance from the 
level of P to the curve, 
when x = x' ; dy, the dis- 
tance from the level of P 
to the tangent. Moreover, the difference 

Ay — dy = eAx 

is shown geometrically as the line QP', and is obviously from 
th& figure an infinitesimal of higher order than Aa; = PM. 

It is also clear from the figure that Ay and dy are equal 
when and only when the curve y = /(a;) is a straight line ; i.e. 




INFINITESIMALS AND DIFFERENTIALS 93 

when f(x) is a linear function, 

f{x) = ax + 6. 

Hitherto x has been taken as the independent variable, Aa; 
as the principal infinitesimal. We come now to the theorem 
on which the whole value of differentials for the purpose of 
performing differentiation depends. 

Theoeem. The relation (4) : 

dy = D,ydx, 

is true, even when x and y are both dependent on a third vari- 
able, t. 

Suppose, namely, that x and y come to us as functions of a 
third variable, t : 

(6) a;=<^(0, 2/ = ^(0, 

and that, when we eliminate t between these two equations, we 
obtain the function „, ^ 

Then dx and dy have the following values, in accordance with 
the above definition, since t, not x, is now the independent 
variable, At the principal infinitesimal : 

dy = Diy At, dx = D^x At. 

We wish to prove that 

dy = Dj/ dx. 
Now by Theorem V of Chap. II, § 6 : 

D,y = Dj/D,x. 
Hence, multiplying through by At, we get : 
D,yAt = D^y- DiXAt, 
or dy = D^dx, q. e.d, 



94 CALCULUS 

With this theorem the explicit use of Theorem V in Chap. 
II, § 5 disappears, Formula V of that theorem now taking on 
the form of an algebraic identity : 

du _ du dy 
dx dydx 

To this fact is due the chief advantage of differentials in the 
technique of differentiation. 

Differentials of Higher Order. It is possible to introduce 
differentials of higher order by a similar definition : 

(7) dH/ = DJh,Aifi, ^y = DJyA3f, etc., 

X being the independent variable. We should then have by (3) 

(8) d^=DJh/dx^ or ^,= ^-'^' etc. 

Unfortunately, however, relation (8) does not contiaue to 

hold when x and y both depend on a third variable, t. For 

example, suppose ^„ , ^„ 

^ ' ^^ x = t% y = a + t\ 

Then y = a + x. - 

When t is taken as the independent variable, we have ac- 
cording to relation (8) : 

d^ = nfydt^ = 2dt'^; 
and since dx = 2tdt, 

it follows that ^ ^ 2dt' ^ J^ ^ J_ . 

dx^~4:Pdfi~2fi~2x' 

On the other hand, when x is taken as the independent vari- 
able, relation (8) becomes 



and consequently ^2^^ 



dh) = D,hfdx^=% 

d^_ 
da?' 



INFINITESIMALS AND DIFFERENTIALS 95 

Thus the quotient, — ^, is seen to have two entirely distinct 

values according as f or a; is taken as the independent variable. 
We will agree, therefore, to discard this definition. The nota- 
tion — ^ as meaning DJ'y is, however, universally used in the 
Calculus, and so we wUl accept the definitions 

^=D^% g=2>.V, etc., 

interpreting the left-hand sides of these equations, however, 
not as ratios, but as a single, homogeneous (and altogether 
clumsy !) notation for that which is expressed more simply by 
Cauchy's D. 

Remark. The operator D^ shall be written when desired as 

—■ Thus 
dx 



X)j. appears as 



d X 



a — X dxa — X 

Again, the equation 
appears as 



d'y_ d dy 
da? dx dx 

Finally, the following notation is sometimes used : 

— = D^y dx, -2 = n sj, ^ etc. 
dx '" ' da;2 '^ ' 

\^i. Technique of Differentiation. Consider, for example, 
Formula II, Chapter II, § 6 : 

On writing this formula in terms of differentials, we have 

d{u + v) _ du . dv 
dx dx dx 



96 CALCULUS 

Now multiply this equation through by dx : 

d(u + v)=du + dv. 

Hence the theorem : 7%e differential of the sum of two ftmctions 
is equal to the sum of the differentials of these functions. 

The others of the General Formulas, Chapter II, §§ 6, 7, 
can be treated in a similar way and lead to corresponding 
theorems in differentials, embodied in the following important 
group of formulas. 

Gbnbkal Formulas of Diffkkbntiation. 
I. d(cu)=cdu. 

II. d{u + v)= du + dv. 

III. d{uv)= udv + vdu. 

-TTT ,M vdu — udv 

V v'- 

As already explained, Theorem V reduces to an obvious 
algebraic identity : w w /7 

(a/Hi CEW WV 

dx dy dx' 

and so does not need to be tabulated. 

Of the special formulas hitherto considered, only two need 
be tabulated, namely : 

Spboial Formulas of Difpbkentiation. 
1. dc = 0. 

The first of these formulas says that the differential of a 
constant is zero. The second is valid, not only when x is the 
independent variable, but when x is any function whatever of 
the independent variable, t. Thus if 

(1) M=vr^ 



INFINITESIMALS AND DIFFERENTIALS 97 

and we set 

(2) x = l-t, 
equation (1) becomes 

(3) u = x^. 
Hence du = -| x~^ dx. 

But dx = dl + d{—t)=0 — dt, 

and thus . -dt du 1 

du = ^^^— or 



2V1 - 1 di 2Vl - 1 

The student should copy off neatly on a card the size of a 
postal the General Formulas I-IV, the Special Formulas 1., 
2., leaving room for a few further special formulas. All the 
difEerentiations of the elementary function of the Calculus are 
based on these two groups of formulas. 

To differentiate a function means henceforth to find either 
its derivative or its differential. Of course, when one of these 
is known, the other can be found by merely multiplying or 
dividing by the difEerential of the independent variable. 

We proceed to show by a few typical examples how differen- 
tials are used in differentiation. 

Example 1. Let m = 12 — Sa; + la?. 

To find du. 

Take the differential of each side of this equation, and apply 
at the same time Formula II : 

du = c?(12)+ d{- 5x)+ d{l3?). 

By Formula 1, d(12) = 0. 

By Formula I, 

d{—5x)=: — 5dx and d(7af^=7da?. 

Hence du = — 5dx + 21ifidx 

= (-5 + 21x^)dx 

and ^ = -6 + 21a;«. 

dx 



98 CALCULUS 

These steps correspond precisely to the steps the student 
would take if he were using derivatives, only he would not 
have written them all out in detail. He would have written 
down at sight : _ f. oh » 

He can avail himself of the facility he has already acquired 
and shorten the work as follows. Since 

du = Dj/idx, 
he can begin by writing 

du=( )dx, 

and then fill in the parenthesis with the derivative.* 

Example 2. Let _ a^ — a? 

To find_dM. 

By Formula IV we have : 

^ ^ (a= + x'^)d(a^ — g') - (g" — x'')dia' + a;') 
(a2 + a;2)2 

^ (a^ + x-'-)(-2x dx) - (g- - x^)(2xdx) 
(a2 + a;2)2 

ia'^xdx 



du 4 a^x 



dx (a2 + xy 

The student would probably prefer to work this example as 
follows. Remembering that 

du = Djjt, dx, 

* The student must be careful not to omit any differentials. ' If one 
term of an equation has a differential as a factor, every term must have 
a differential as a factor. Such an equation as 

dM = — 5 + 21 x2 
is absurd, since the left-hand side is an infinitesimal and the right-hand 
not. Moreover, there is no such thing as djU. 



INFINITESIMALS AND DIFFERENTIALS 99 

iDegiii by writing 

du = dx, 

and then fill in the fraction by the old familiar methods of 
Chapter II. 

In the two examples just considered, the processes with 
difEerentials correspond precisely to those with derivatives^ 
with which the student is already familiar. , This will always 
be true in any differentiation in which composite functions are 
not involved ; i.e. whenever, according to our earlier methods, 
the vanished Theorem V of Chapter II, § 8 was not used. It 
is in the diEEerentiation of composite functions that the method 
of differentials presents advantages over the earlier method. 
We turn in the next paragraphs to such examples. 

EXERCISES 

Differentiate each of the following functions by the method 
of differentials, and test the result by the methods of Chap- 
ter II. 

i^"!. u = x^ — Zx + 1. 

2. y ^=a+bx + ex'': 

3. w = a' — »'. 

4. s = 96<-16«2. 



5. s = vat + ^gt\ 

1 — x A T —2dx 

6. M = - Ans. du = - 



Ans. du : 


= 3xMx 


-,3dx. 


Ans. dy = 


= bdx +2cxdx. 


Ans. 


dw = — 


Sz^dz. 


Ans. 


^=96 
dt 


-32t. 


■Ans. ^=vo + gt. 
dt 



1 + x (1 + xy 

'^ Ans. dy^^"-'"'^" 



" 1+x^ " (l-l-a;2)2 

1 + x + x^ . ,a;2 — 1, 

8. g= ^ ^ — • .4ms. dz = —- dx. 

2x 2x^ 

3-2x+a? ,„ n* — ^i 

9. u = - , , • 10. y = - 



4: + x^ — a? a^ + a'^x^ + X^ 



100 CALCULUS 

5. Continuation. Differentiation of Composite Fanctions. 



Example 3. Let u = Vl + x+xK 

To find — . 
dx 

Here, we begin by computing du. To do this, introduce a 
new variable, y, setting 

2/ = 1 + a; + a;'. 

Then u = y^. 

Next, take the differential of each side of this equation. By- 
Special Formula 2 above, 

du = dy^= \y~^ dy. 

Moreover, dy = {1 + 2 x)dx. 

Hence d« = Jl±l^^ 

2V1 -\- x+x^ 



and- 



du _ 1 + 20! 



dx 2-Vl+x + x^ 



Let the student carry through the above differentiation by 
the methods of Chapter II and compare his work step by step 
with the foregoing. He will find that, although the two 
methods are in substance the same," the method of differentials 
is simpler in form, since no explicit use of Theorem V here is 
made. 

Abbreviated Method.* The solution by differentials can be 
still further abbreviated by not introducing explicitly a new 

* The student should not hasten to take this step himself. He will do 
well to omit the text that follows till he has worked a score or more of 
problems in differentiating composite functions as set forth under Ex- 
ample 3, introducing each time explicitly a new variable, as y, z, etc. 
Not until he comes himself to feel that the abbreviation is an aid, should 
he attempt to use it. 



INFINITESIMALS AND DIFFERENTIALS 101 

variable, y. The problem is to find du, when 

u=(l + x + a:2)i 

Now, Special Formula 2, as has already been pointed out, 
holds, not merely when x is the independent variable, but for 
any function whatsoever. It might, for example, equally well 
be written in the form : 

d [<^(a;)]" = ra[^(a;)]»-id<^(a!). 

In the present ease, then, the content of that theorem, — the 
essential and complete truth it contains, — enables us to write 
down at once the equation : 

- d{l + x + x^)^= 1(1 + a; -I- a;^)"* d(l + x + x^). 

This last differential is computed at sight, and thus the 
answer is obtained in two steps. 

Even these two steps are carried out mentally as a single 
process, when the student has reached the highest point in the 
technique of differentiation. He then thinks of the formula : 

J r dx 
d-\/x = - 



2y/x 



realizes that it holds, not merely when x is the independent 
variable, but for any function of x, and so writes down first 
the easy part of the right-hand side of the equation, thus : 



d-\/l + a; + a;2 = 



2V1 + a; + a;2' 



carrying in his head the fact that the numerator is the differen- 
tial of the radicand, i.e. d(l + x + x^). This differentiation he 
performs mentally, and thus has the final answer with no 
intermediate work on paper : 



102 CALCULUS 

Example 4. The method of differentials is especially use- 
ful in the case of implicit functions. Thus, to find the deriva- 
tive of y with respect to x when 

a? — ^xy-\-2y* — l. 

Take the differential of each side : 

^x^dx - 3xdy - 3ydx + 8fdy = 0. 

Next, collect the terms in dx by themselves ; the others will 
contain dy as a factor : 

(3x^ - 3y)dx + (8f - 3x)dy = 0. 

Hence ^^3_y-3^, 

dx sy' — 3a! 

EXERCISES 

Differentiate the following twelve functions by the method 
of differentials and also by the methods of Chapter II (in 
either order), introducing each time explicitly the auxiliary 
variable, if one is used. 

1/ 1. u=V a^ + a^^ + a^. Ans. du = J^^±^^^- 

Va* -f- aV + !B* 

T 3- y = — • Ans. dy = 






3. M = Ans. du = ■ 



dx 



J " 3. u = - j3.ns. au = — -• 

T -y 1-x (l-®)' 

A If Suggestion. Introduce an auxiliary variable y = l — x. 



\ Then u = i 
/V 1^ 4. u=— — - 



Ans. — - 



\ -, " ' (1 - xy dx (1 - x)' 

I v^ , 1 . dy — 2x 

v/ ■^ H-as" dx (l + x^y 



INFINITESIMALS AND DIFFERENTIALS 103 



6. s = — • Ans. — = — - 



(a + ty y dt (a 4 ty 

7. 2x''-xy + iy' = 5. Ans. ^ = ^'"~^ . 

dx x — Sy 

8. a!2/ = al ^ns. ^ = -1. 

dx X 

9. 2/2 = 2ma;. Ans. ^=^. 

da; 2/ 

10. ^ + ^=L ^ns. ^ = -^. 

11. 2a!2 + 3y2 = 10. ^ns. ^ = _?^. 

da; 3y 

12. 2a!2/-a; + !/ = 0. Ans. ^ = 3lJi1|. 

da 2a! 4-1 

The student can work the problems at the end of Chapter 
II by the method of differentials. Tor further practice, if de- 
sired, the following examples are appended. 

du la^—lx"- — ! 



13. u={x^+l)y/3?-x. Ans. — = 



dx 2 Va;' ■ 



• X 



14. y={x + 2V)(x-hY. Ans. ^ = 3(x^-b^). 

dx 



15. u = - 



Va^ - x^ 
16. u = yj: 



la — X 

<l— 



17. u=- 



X 

X — a 



Ar, 


,.s. *i= 
dx 


a2 






V(a^- 


xy 


Ans. 


du_ 
dx 


a 






2x^ax- 


— x' 


Ans. 


du 


d? 





^2ax — x^ dx V(2aa; — 0:2)3 

18. u = f^l±^\ ■ Ans. *f=2- »^ 



L^^Y 



a; / da; a;* — a^ 

19. ,=(t±b^y. Ans. ^=42^i^^ 

\ y^ J dy 2/5 



104 CALCULUS 

20. u = — — Va" — a?. Ans. — ■= ■ 

a? cfa; af Va'- — a? 



„, ja^— x-\-l . du 

21. w=-\/ ■ — • Ans. — = 

^x'' + x + l dx 



3a* 



+ a; + 1 dx (3-2+ X +. 1) Va^ + x^ + l 



23. u 



__ (x — a^)' . du _ S-Vx — x^ 

~ x^ ' dx 3a^ 

= (xi-aiy. Ans. *^ = iMzi^. 



dx 3a,f 

du 



24. u = x(x^ + 5)\ Ans. ^ = 5(»' + l)(a!3 + 5)i 

dx 

25. a;' + y^ = a'. -4ns. -^ = — \;'" . 

da; *a; 



CHAPTER V 
TRIGONOMETRIC FUNCTIONS 

1. Radian Measure. In Trigonometry, the radian measure 
of an angle was introduced, apparently for no good purpose. 
The reason lies in the importance for the Calculus of this new 
system of measurement, and will become clear in the next 
paragraph, when we come to differentiate the sine. We will 
first recall the definition. 

Let a circle be described with its centre at the vertex O of 
the angle ; let r denote the length of the radius of the circle 
and s, that of the intercepted arc. Then 
the radian measure, 6, of the angle is 
defined as the ratio s/r : 



(1) 




For a right angle, s = — - , and hence 6 ■■ 



Fig. 34 

A straight angle 



has the measure 



61 = T = 3.14159 26535 89793 • - .. 



Let <l> be the measure of the given angle in degrees. Then 
d and <j> are proportional, 

e = c<J3, 

where c is a constant. To determine c, use a convenient angle 
whose measure is known in both systems ; for example, a 
straight angle. For the latter, 



and 
105 



<l> = 180. 



106 CALCULUS 

Substitating these values in the above equation we find : 



and hence 


7r = cl80, 


''"180' 


(2) 


180 '^' 


TT 


This equation can also be written in the form 


(3) 


TT 


180 



and thus an easUy remembered rule of conversion from radian 

measure to degree measure, or the opposite, obtained: The 

radian measure of an angle is to w as its degree measure is to 180. 

The unit of angle in radian measure, i.e. the angle for which 

^ = 1 and hence s= r, 

is called the radian. It is obvious geometrically that it is a 
little less than 60°. Its precise value (to hundredths of a sec- 
ond) is given by (2) : 

<!> \^, = ^ = 57° 17' 44.81" (= 57.29578°). 

TT 

On the other hand, the radian measure of an angle of 1° is 

1 = J!L = .01745 32925 19943 -. 

'*-' 180 

The student should practice expressing the more important 
angles, as 30°, 46°, 60°, 90°, 120°, etc., in radian measure until 
he is thoroughly familiar with the new representation for 
them. 

If, in particular, the radius of the circle is taken as Tinity, 
then 6 and s are the same number : 

(4) 6 = s, when r = 1 ; 

or the arc is equal to the angle. Thus the radian measure of an 
angle might have been defined as the length of the intercepted 



TRIGONOMETRIC FUNCTIONS 



107 



are in the unit circle (i.e. the circle of unit radius with its 
centre at 0). 

Graph of sin x. It is important for the student to make an 
accurately drawn graph of the function 

y = siax, 

X being taken in radian measure. Let the unit of length, as 
usual, be the same on both axes, and let it be chosen as 1 cm. 
For this purpose Peirce's Table of Integrals (the table of 
Trigonometric Functions near the end) is especially convenient, 
since the outside column gives the angles in radian measure, 
and thus as many points of the graph as are desired can be 
plotted directly from the tables. 



j/=sin X 




Fig. 35 

Since sin (tt — a;) = sin x 

each determination of the coordinates {x, y) of a point on the 

graph, for which < a; < ^ yields at once a second point, 

namely (tt — x, y). Thus one arch of the curve is readily con- 
structed from the Tables.* 

From this arch a templet, or curved ruler, is made as fol- 
lows. Lay a card under the arch and with a needle prick 
through enough points so that the templet can be cut ac- 
curately with the scissors. 

By means of the templet further arches can be drawn me- 
chanically, and thus the curve is readily continued in both 

* The graph could be made directly without tables from purely geomet- 
rical considerations. Draw a circle of unit radius. Construct geomet^ 
rically convenient angles, as those obtained from a right angle by 
successive bisectors. Measure any one of these angles, X ABP„, in ra^ 
dians and this number will be the abscissa of the point on the graph, the 



108 



CALCULUS 



directions to the edges of the paper.* Put this curve in the 

upper quarter of a sheet of centimetre paper. 

The graph brings out clearly the property of the function 

expressed by the word periodic. The function admits the 

period 2ir, since 

sm (a; + 2 tt) = sin a; 

Graph of cos x. By means of the templet the graph of the 

function 

y = cos X 

can now be drawn mechanically. This function also admits 

the period 27r : / „ ^ 

cos (a; + 2 tt) = cos x. 




ordinate being the perpendicular dropped from P„ on tlie line BA. Thus, 
if n = 3, the coordinates of tlie point on the graph are : 




= ^ = 1.18, 



y = .92. 



A second point of the arch, 
that corresponding to Pj, has 
the same y, its coordinate be- 
ing 

x = 7r-^=1.96, y=.92. 
8 

Of course, the distance ir must be laid off on the axis of x by measui'e- 
ment ; it cannot be constructed geometrically from the unit length. This 
done, the further abscissae are found by successive bisectors. 

* In order to obtain the most satisfactory figure, observe that the 
curve has a point of inflection at each of its intersections with the axis of 
X, the tangent there making an angle of ± 45° with that axis. Since a 
curve separates very slowly from an inflectional tangent, it will be well to 
draw these tangents vrith a ruler. On laying dovro the templet, the curve 
can then be ruled in from the latter vrith great accuracy. It will not 
separate sensibly from its tangent for a considerable distance from a 
point of inflection. 



TRIGONOMETRIC FUNCTIONS 



109 



Put the graph in the second quarter of the sheet, choosing the 
axis of y for this curve in the same vertical line as the axis of 
y for the sine curve above. There remains the lower half of 
the sheet for the next graph. 



Graph of tan x. The same tables make it easy to plot points 
profusely on the graph of the function 

y = tan x 

in the interval 0^ a: < -. Take the axis of y in the same ver- 

i I I I 

2/=tan± 



¥ 



^ 



Fig. 38 

tical line as in the case of the preceding graphs. This done, a 
second templet is made and by means of it the graph is drawn 

mechanically for values of x such that — ^ < a; < 0. 

It is desirable furthermore to plot the function in the two 
adjacent intervals 



no 



CALCULUS 



"■ ^ « ^ Stt 



-T<^<-2' 



in order to suggest the fact that this function admits the 

period it: , , ^ . 

tan {x + ir) — tan x. 

2. Differentiation of sin x. To differentiate the function 

(1) y = smx, 

apply the definition of a derivative given iu Chap. II, § 1. 

Give to X an arbitrary value 
cBb and compute the corre- 
sponding value y^oiy; 

yo = sin Xo- 

Then give x an increment Ax, 
and compute again the corre- 
sponding value of y : 

yo + Ay = sia (xa + Ax). 

Hence 

Ay = sin (ajo + Ax) — sin Xq, 




Fig. 39 



M' M 



(2) 



Ay _ sin (xp + Ax) — sin a;,. 
Ax Ax 



It is at this poiat in the process that the specific properties 
of the function sin a; come into play. Here, the representar 
tion of sin a; by means of the unit circle, familiar from the 
beginning of Trigonometry, is the key to the solution. From 
the figure it is clear that 

siu Xo = MP, sia {x^ + Ax) = M'P', 

Ay = sin (xq + Ax) — sin aio = QP', Ax = PP'. 



Hence 
(3) 



Ax ppi' 



TRIGONOMETRIC FUNCTIONS 111 

and so we want to know the limit approached by the latter 

limS^. 

By virtue of the Fundamental Theorem of Chap. IV, § 2, we 
can replace this ratio by a simpler one, since the arc PP and 
the chord PP' are equivalent infinitesimals : * 

lim5 = l. 
PP 

Hence lim ^' = lim ^. 

p'=p jypf F'=p ppt 

On the other hand, the triangle QPP' is a triangle of refer- 
ence for the 2^ QPP = <f>, and so 

■ ■ ^ — = sin <A. 
PP 

When P approaches P, the secant PP' (i.e. the indefinite line 
determined by the two points P and P) approaches the tan- 
gent PT at P, and thus 

lim <i> = -4. qPT =1-01^ 
p'=p J 

Finally, then, 

lim ^ — = lim sin <^ = sinf ^ — Xq )= cos Xq, 
P'=p PP p'=p \2 J 

and consequently j- ^^ A^ _ ^^^ ^^^ 

A»a) Aa; 

* The student should assure himself of the truth of this statement by 
visualizing the figure (making an accurate drawing with ruler and com- 
pass for angles of 30°, 15°, and 7J°, the circle used being 10 in. in 
diameter) and realizing that, when P' is near P, the difference in length 
between the arc and the chord is but a minute per cent of the length of 
either one. A formal proof will be found below. 



112 CALCULUS 

or, on dropping the subscript, 

(4) DjSin X = cos x. 

This theorem gives rise to the following theorem in dif- 
ferentials : 

(5) d sin X = cos x dx. 

Reason for the Radian. The reason for measuring angles in 
terms of the radian as the unit now becomes clear. Had we 
used the degree, the increment \x would not have been equal 
to PP ; we should have had : 

Aa; PP' . 180 ^^, 
= , or Aa; = PP\ 

360 2 TT IT 

Hence (3) would have read : 

Ay_ TT QP 
Ax 180 ppi' 

and thus the formula of differentiation would have become : 

D^ sin X = -^ cos x. 
180 

The saving of labor in not being obliged to multiply by this 
constant each time we differentiate is great. Still more impor- 
tant, however, is the elimination of a multiplier which is of 
the nature of an extraneous constant, whose presence would 
have obscured the essential simplicity of the formulas of the 
Calculus. 

EXERCISE 
Prove in a similar manner that 

D^ cos x = — sin x. 

3. Certain Limits. In the foregoing paragraph we have made 
use of the fact that the ratio of the arc to the chord approaches 
1 as its limit. A. formal proof of this theorem, based on the 




TRIGONOMETRIC FUNCTIONS 113 

axioms of geometry, can be given as follows. Draw the tan- 
gent at P and erect a perpendicular at P' cutting the tangent 
iu Q. Denote the angle 2(! P'PQ by a. 

Then . ^ 

PP <PP' <Pq + FQ; 

P 

for i) a straight line is the shortest dis- 
tance between two points ; and ii) a convex curved line is less 
than a convex broken line which envelops it and has the same 
extremities. But 

PQ = EJ^, P'Q = PPta.Tia. 

cos a 

Hence J!P' 1 

1 < < 1- tan a. 

PP cos a 

When « approaches 0, the right-hand member of the double 
inequality approaches 1 ; hence the middle member must also 
approach 1, or -^ 

lim = 1, q. e. d. 

PP 

The foregoing proof holds, not merely for a circle, but for 
any curve with a convex arc PP'. Consequently the theorem 
is established generally. 

The Limit lim ^^^. From Pig. 41 ^"^ 



it is clear that 




- MA 

JfP=sin«, AP=a, Fig. 41 



and hence 



sin a _ MP 
« " AP 



By direct inspection of the figure it is seen, then, that 
.(1) lim?i^=l. 



114 



CALCULUS 



A formal proof of this equation can be given as foUowa 

From Fig. 42 ^ 

PP' = 2sina, PP' = 2a. 



Hence 



sina_PP' 



and therefore, by the proposition just established, 



lim?^ = lim^'=l. 
0=^0 a p'=pppi 



Another Proof of (1). The area of the sector OAP, Fig. 42, 
is \ a, and it obviously lies between the areas of the triangles 
OMP and OPN. Hence 

^ sin a. cos « < ^ a < ^ tan a 




or 



cos a < -^— < 



sm a cos a 



•jpr When « approaches 0, each of the ex- 
treme terms approaches 1, and so the 
middle term must also do so, q. e. d. 
From Peirce's Tables, p. 130, we see that 

sin 4° 40' = .0814, 

and the same angle, measured in radians, also has the value 
.0814, to three significant figures. Thus for values of a not 
exceeding .0814, sin a differs from a by less than one part in 
800, or one-eighth of one per cent. 



The Limits lim ^ - "os « ^^^ lim ^~'^°^" - From Fig. 42, 
the first of these limits is seen to have the value : 



lim- 



cos« 



= 0. 



TRIGONOMETRIC FUNCTIONS 



115 



A formal proof can be derived at once by the method em- 
ployed in the evaluation of the next limit, 

1 



lim- 



cos a 



Expressing 1 — cos a. in terms of the half angle, we have 
l-cos« = 2sin2^. 



Hence 



and 



1 — cos a 



Ssin^? 



2 1 
''2 



lim 



l-cos« _li,^ 

«2 2 -^ 



. Of 

a 
2 . 



EXERCISES 

In the accompanying figure determine the following limits 
when ot approaches : 



1. lim 



2. lim 



3. lim 



AR 
MP 
AQ 
AP 
BQ 
MP 



Ans. — 
2 

Ans. 1. 
4. lim 




FlQ. 43 



6. lim — — • 



7. lim 



HP 
AP 

PQ 



AN 



6. lim 



8. lim 



PN 
AP 
BQ 

pn' 



Determiae the principal part of each of the following infini- 
tesimals, referred to a as principal infinitesimal : 

9. MP. Ans. a. 10. PB. Ans.^a. 11. BQ. 

12. PN. 13. AQ. 

15. PQ. 16. MN. 



14. MA. Ans. I a'. 
17. AQ-MP. 



116 CALCULUS 

4. Critique of the Poregoiag Differentiation. The differenti- 
ation of sin a; as given in § 1 has the advantage of being direct 
and lucid, and thus easily remembered. Each analytic step is 
mirrored in a simple geometric construction. It has the dis- 
advantage, however, of incompleteness. For, first, we have 
allowed Ax, in approaching 0, to pass only through positive 
values ; and secondly we have assumed Xg to lie between and 
^ IT. Hence there are in all seven more cases to consider. 

An analytic method that is simple and at the same time 
general is the following. Eecall the Addition Theorem for 
the sine : 

sin (a +b)= sin a cos b + cos a sin b, 

sin (a — 6) = sin a cos 6 — cos a sin 6, 

whence sin (a -(- 6) — sin (a — &) = 2 cos a sin h. 

Let a + b = XQ+ Ax, a — b = XQ. , 

Solving these last equations for a and b, we get : 

, Ax , Ax 

and the difference-quotient becomes 

. Ax 
sm — 
Aw / Ax\ 2 

-=cos(^a.o+-^j-^- 

2 

The first factor on the right approaches the limit cossjo 
when Aa; approaches 0. On setting ^Ax= a, the second fac- 
tor becomes 

sma 



Hence the factor approaches 1. Thus 

lim — = cos Xq, 

ctx^Ax 



TRIGONOMETRIC FUNCTIONSv 117 

or, on dropping the subscript, 

Z)j. sin X = cos X. 

5. Differentiation of eosa;, tanw, etc. To differentiate the 
function cos x, introduce a new variable, y, by the equation 

• y=^—x. Hence x = - — y, 

and cos x = cos ( — - y]= sin y. 

Taking the differential of each side of the equation thus ob- 
tained, we have : 

dcosa; = dsmy = oos ydy. 



But 


cosy = 


sin X and dy = 


-dx. 


Hence 






1 


(1) 




d cos aj = — sin a; da;. 




To differentiate the function tan x, set 








.„„ sin a; 
tan x = 








cos a; 




Hence 


dtau X 


cosa;dsina; — sina;d 

COS^ X 


cos a; 






_ cos" a; da; 4- sin^ a;da; _ 


dx 


and thus 




cos^a 


cos^a;' 


(2) 




dtana;— sec^xda;. 





It is shown in a similar manner (or by setting x= — — y in 
the equation just deduced) that 

(3) d cot a; = — cs c^ a; da;. 

These are the important formulas of differentiation for T;he 
trigonometric functions. By means of them all other differen- 
tiations of these functions can be readily performed. Thus, 



118 CALCULUS 

to difEerentiate the function sec x, set 
sec X = (cos a;)~*- 

Then dsecx = = . 

cos^ X cos^ X 

It is not desirable to tabulate the result, since one rarely 
has occasion to difEerentiate either sec a; or csco;, and when 'the 
occasion does arise, the difEerentiation can be worked out 
directly, as above. 

The student should now add to his card of Special Formulas 
the four main formulas just obtained. This card will now read 
as follows : 

1. dc = 0. 

2. - dx" = ma;"~i dx. 

3. d sin X = cos x dx. 

4. d cos x = — sin x dx. 

5. . d tan x = sec" x dx. 

6. d cot x = — esc" X dx. 

6. Shop Work. To acquire facility in the use of the new 
results, the student should work a generous number of simple 
examples, for which the following are typical. 



Example 


1. To differentiate the function 






M= sin aa;. 


Let 




y = ax. 


Then 




M = sin y. 


and 




du = daiay = cosydy. 


But 




dy = a dx. 


Hence, 


substituting, we have 




du 


= acoaaxdx or — sin aa; = a cos ax. 
dx 



TRIGONOMETRIC FUNCTIONS 119 

The solution can be abbreviated as follows. The equation 

d sin X = cos x dx 

is true, not merely when x is the independent variable. It 
holds, for example, in the form 

dsmy = COS ydy, 

where y is any function of x. Hence we can write immediately 

dsin ax = cos ax d{ax), 

and thus obtain the result 

d sin ax^a cos ax dx. 

Example 2. To differentiate the function 



M=V1 — fc^sin",^. 
Let z = 1 — fc2 sin" <^. 

Then u = z^ 

du = dz' = ^z~^dz; 

dz = -K'dairs?^. 

Let y = sin <j}. 

Then dy = cos tj> d^ 

and dsin' </i = dy^ = 23/% = 2sin<^ cos <^d<^. 

Hence c?m=^«"^(— 2fc^sin<^ cos (^d<^) 

du fc2 sin d cos A 
or — = ^ -!— . 

d<l> Vl — A;2 sin2 </. 

Example 3. If siax + 3my = x- y, 

to find -^. Take the differential of each side of the equation: 
dx 

cos xdx + cos ydy = dx — dy. 

Hence (cos x — V)dx + (cos y + l)dy = 

, dy _ 1 — cos X 

dx 1 + cos 3^ 



120 CALCULUS 

EXERCISES 
Differentiate the following functions. 

1. M = cos oa;. 

2. y = cos^x. 

3. y = CSC X. 

X 

4. M = tan— ■ 

2 

5. M = cot2a;. 

6. M = sec3a;. 
8. u = sin' X. 

10. u = x + tan X. 
12. M = sec^ X. 
sin a; 



14. w = - 



1 — cos X 



15. u = Vl + cos X. 
1 — cos a; 



16. u = 
18. M = 



1 + cos a; 
sin a; 



20. u = 



a + b cos X 
1 



sin a; + cos x 
22.* M = vers X. 

23.* It = covers x. 



* The versed sine and the coversed siTie are defined as follows : 
vers a; = 1 — cos a; ; covers a; = 1 — sin a. 





dx 


— a sin ax. 




dx 


— 2 sin a; cos a;. 




dy _ 
dx 


esc' x cos X. 




du _ 
dx 


2 2 




du_ 


-2csc2 2a!. 


7, 


. M = 


tan^ ax. 


9 


. U = 


1 — sin X. 


11 


. u = 


cos' X. 


13 


. u = 


sin X cos X. 




du_ 
dx 


_icsc25. 
2- 2 




du_ 
dx 


: 1 sinf. 
a/2 2 


17 


. u = 


1 + sin a; 
1 — sin a; 


19 


. u = 


1 


a cos a; + 6 sin a; 


21 


. u = 


1 


(a + b cos a;)^ 




du 
dx~ 


: sin X. 




du_ 
dx 


— cos a;. 



TRIGONOMETRIC FUNCTIONS 121 



cos? 



24. u = x sin. 2x. 25. m = 



X 



28. M = taii-^- 29. u = 5iBJ!^. 

1 -^ X X 

30. M = sin a; + cos 2 a;. 31. m = a;^ cos tts;. 

1 «„ cos d> 

32. M=- 33. M = — z=:^===- 

Vl-A!''sin2^ Vl — &2 sini* <^ 

■ / . \ (^V COS (a; + w) — cos 11 

34. a!COS«= sm(!B + w). ^^ = -, '- ^ "^ r-^- 

dx sm (a; + y) + a; sm y 

35. tana;— cot 2/= sin a; sin 3/. 36. sin x + sin jr = 1. 
37. tan + tan <^ = 2 tan <^ tan 6. 38. a; = 2/ sin y. 

7. Maxima and Minima. By means of the new functions 
studied in this chapter the range of problems in maxima and 
minima which can be treated by the Calculus has been materi- 
ally enlarged. No new principles are involved ; the student 
should go over carefully the paragraphs of Chap. Ill relating 
to this subject, before he proceeds farther with the present 
paragraph. 

Example 1. A man in a rowboat 1 mile off shore wishes to 
go to a point which is 2 miles inland and 4 miles up the 
beach. If he can row at the rate of 5 miles an hour, but can 
walk only 3 miles an hour after he lands, in what direction 
should he row in order to get to his destination in the shortest 
possible time ? 

In the first place, it is clear that the straight line AEB is , 
not the best path. For, if he rows toward a point P slightly 
farther up the beach, the amount by which he lengthens the 
leg AP of his path is very nearly equal to the amount by which 



122 



CALCULUS 



he shortens the leg PB.* Consequently the time is short- 
ened. 

On the other hand, P obviously ought not to be taken so 

far up the beach as D. 
The minimum occurs, 
therefore, for some in- 
termediate point. 

Let the angles 0, <f> 
be taken as indicated 
in the figure. Then, 

since i = -, 
Fig. 44 V 

time from A to P = ^^^ = — - — ; 
5 6 cos 6 

time from P to 5=:^= ^ 







^'7 

X/ 


B 
2 


X 






sy 


V 






'' ^-<r 


P D 


1 






A 









3 cos (9 

Hence the total time, u, which is to be made a minimum is 

1 . 2 



(1) 



; + ; 



5 cos 6 3cos ^ 

Moreover, 6 and <^ are connected with each other by a rela- 
tion which is readily obtained by expressing the distance CD 
in two ways : 

(2) tane +2tan<^ = 4. 

We are now ready to compute du/d6 and set it equal to : 

J _ _ dcos^ _ 2 d cos <i> 
~ 5 eos^ 3 eos2 ^ 



(3) 



sec^ sin 6 ■,„ , 2 sec'' tb sin <A , . 
= g (w-l ^ ^dcj}-, 

du _ sec' 6 sin 2 sec' <f> sin <^ d<j) 
d6~ 5 3 W 



* Let the student not leave this statement till he is absolutely con- 
vinced of its truth. An accurate figure on a large scale will bring the 
fact out clearly. 



TRIGONOMETRIC FUNCTIONS 123 

On setting du/dO = 0, we obtain the equation : 
,.-. setfi 6 ajg 6 _ _ 2 sec' <^ sin <^ d^ 



5 Z dd 

Next, differentiate (2) : 

sec'= Odd + 2 Bec2 ^ d</, = 0, 



(5) sec25l = -2sec2^^. 



Now, divide equation (4) by equation (5) : * 

,a\ sin 6 gin <A sin 6 5 

(o) • =i^ — 2: or = - • 

^ '. 5/3 sinc^ 3 



The result, stated in words, is as follows : sia 6 is to Sia <^ as 
the velocity in water is to the velocity on land. 

Let the student work the general . problem, in which all the 
data are take^ in literal form, and verify the general result 
just stated. 

In order actually to determine 9, equations (2) and (6) must 
be solved as simultaneous : 

,„^ I tan 6 + 2 tan <^ = 4, 

^ ' . l3sine = 5sin<^. 

This is done best by the method of Trial and Error, as it is 
called in Physics ; Successive Approximations being the name 
usually given to it in Mathematics. It is a most important 
method in both sciences, and the student should let no oppor- 
tunity go by to use the method whenever, as here, he meets a 
case which calls for it. Cf. Chap. VII, § 5. 

The Corresponding Problem in Optics. We have stated and 
solved a problem which is not lacking in interest, but which 
appears to have no scientiiic importance. This very problem, 
however, occurs in Optics. The velocity of light is different id 

* i.e. divide the left-hand side of (4) by the left-hand side of (5) for a 
new left-hand side ; and do the same thing for the right-hand sides. 



124 CALCULUS 

different media, such as air and water. Suppose two media to 
be in contact with each other, the common boundary being a 
plane. Let ^ be a luminous point, from which rays emanate 
in all directions. When the rays strike the bounding surface, 
they are all refracted and enter the second medium in case the 
velocity of light in that medium is less than in the first. One 
of the refracted rays will pass throttgh a given point B. And 
now the law of light is that the time required for the light to 
pass from ^ to B is less for this path than for any other 
possible path. 

If, then, the velocity of light in the first medium is m * and 
in the second medium, v, we have : 

sin 6 u 
sin<^ V 

where n is the iTidex of refraction for the passage from the first 
medium into the second. 

EXERCISES 

1. A wall 27 ft. high is 64 ft. from a house. Find the length 
of the shortest ladder that will reach the house if one end 
rests on the ground outside the wall. 

Take the angle which the ladder makes with the horizontal 
as the independent variable. 

2. The equal sides of an isosceles triangles are each 8 in. 
long, the base being variable. Show that the triangle of 
maximum area is the one which has a right angle. 

Take one of the base angles as the independent variable, ^. 

3. A gutter is to be made out of a long strip of copper 
9 in. wide by bending the strip along two lines parallel to the 
edges and distant respectively 3 in. from an edge. Thus the 
cross-section will be a broken line, made up of three straight 
lines, each 3 in. long. How wide should the gutter be a;t the 

* The letter u used here has nothing to do with the v, used above in 
solving the problem. 



TRIGONOMETRIC FUNCTIONS 125 

top, in order that its carrying capacity may be as great as 
possible ? Ans. 6 ia. 

4. Johnny is to have a piece of pie, the perimeter of which 
is to be 12 in. If Johnny may choose the plate on which the 
pie is to be baked, what size plate would he naturally select ? 

5. A can-buoy in the form of a double cone is to be made 
from two equal circular iron plates by cutting out a sector 
from each plate and bending up the plate. If the radius of 
each plate is a, find the radius of the base of the cone when 
the buoy is as large as possible. Ans. aVf. 

6. From a circular piece of filter paper a sector is to be cut 
and then bent into the form of a cone of revolution. Show 
that the largest cone will be obtained if the angle of the sector 
is .8165 of four right angles. 

7. Two solid spheres, whose diameters are 8 in. and 18 in., 
have their centres 35 in. apart. At what point in their line 
of centres and between the spheres should a light be placed in 
order to illuminate the largest amount of spherical surface ? 

Ans. 8 in. from the centre of the smaller sphere. 

8. Find the most economical proportions for a conical tent. 

9. A block of stone is to be drawn along the fioor by a rope. 
Find the angle which the rope should make with the horizontal 
in order that the tension may be as small as possible. 

Ans. The angle of friction. 

10. A block of stone is to be drawn up an inclined plane by 
a rope. Find the angle which the rope should make with the 
plane, in order that the tension in the rope be as small as 
possible. 

11. A statue ten feet high stands on a pedestal that is 50 ft. 
high. How far ought a man whose eyes are 5 ft. above the 
ground to stand from the pedestal in order that the statue may 
subtend the greatest possible angle ? 

12. A steel girder 25 ft. long is moved on rollers along a 
passageway 12.8 ft. wide, and into a corridor at right angles 



126 CALCULUS 

to the passageway. Neglecting the horizontal width of the 
girder, find how wide the corridor must be in order that the 
girder may go round the corner. Ans. 5.4 ft. 

13. A gutter whose cross-section is an arc of a circle is to be 
made by bending into shape a strip of copper. If the width 
of the strip is a, find the radius of the cross-section when the 
carrying capacity of the gutter is a maximum. Ans. a/ir. 

14. A long strip of paper 8 in. wide is cut off square at one 
end. A corner of this end is folded over on to the opposite 
side, thus forming a triangle. Find the area of the smallest 
triangle that can thus be formed. 

15. In the preceding question, when will the length of the 
crease be a minimum ? 

16. The captain of a man-of-war saw, one dark night, a 
privateersman crossing his path at right angles and at a 
distance ahead of c miles. The privateersman was making 
a miles an hour, while the man-of-war could make only b miles 
in the same time. The captain's only hope was to cross the 
track of the privateersman at as short a distance as possible 
under his stem, and to disable him by one or two well-directed 
shots ; so the ship's lights were put out and her course altered 
in accordance with this plan. Show that the man-of-war 

crossed the privateersman's track - ■Va!' — b^ miles astern of 
the latter. 

If a = 6, this result is absurd. Explain. 

17. The illumination of a small plane surface by a luminous 
point is proportional to the cosine of the angle between the 
rays of light and the normal to the surface, and inversely pro- 
portional to the square of the distance of the luminous point 
from the surface. At what height on the wall should an arc 
light be placed in order to light most brightly a portion of 
the floor a ft. distant from the wall ? 

Ans. About ^^ a ft. above the floor. 



TRIGONOMETRIC FUNCTIONS 127 

18. A town A situated on a straight river, and another town 
B, a miles farther down the river and 6 miles back from the 
river, are to be supplied with water from the river pumped 
by a single station. The main from the waterworks to A 
will cost $ m per mile and the main to B will cost % n per mile. 
Where on the river-bank ought the pumps to be placed ? 

19. A telegraph pole 25 ft. high is to be braced by a stay 
20 ft. long, one end of the stay being fastened to the pole and 
the other end to a short stake driven into the ground. How 
far from the pole should the stake be located, ia order that the 
stay be most effective ? 

20. Into a full conical wine-glass whose depth is a and 

generating angle « there is carefully dropped a spherical ball 

of such a size as to cause the greatest overflow. Show that the 

radius of the ball is 

a.sm a 



sin a + cos 2 a 



21. A foot-ball iield 2 a ft. long and 2 b ft. broad is to be 
surrounded by a runnhig track consisting of two straight sides 
(parallel to the length of the field) joined by semicircular ends. 
The track is to be 4 c ft. long. Show how it should be made 
ia order that the shortest distance between the track and the 
foot-ball field may be as great as possible. 

22.* The number of ems (or the number of sq. cms. of text) 
on this page and the breadths of the margius being given, 
what ought the length and breadth of the page to be that the 
amount of paper used may be as small as possible ? 

23. Assuming that the values of diamonds are proportional, 
other things being equal, to the squares of their weights, and 
that a certain diamond which weighs one carat is worth $ m, 
show that it is safe to pay at least $ 8 m for two diamonds 
which together weigh 4 carats, if they are of the same quality 
as the one mentioned. 

* Exs. 22-25 do not involve Trigonometry. 



128 



CALCULUS 



24. When a voltaic battery of given electromotive force 
(E volts) and given internal resistance (r ohms) is used to 
send a steady current through an external circuit of R ohms 
resistance, an amoimt of work, W, equivalent to 

^^ X 10' ergs 

is done each second in the outside circuit. Show that, if dif- 
ferent values be given to M, W will be a maximum when 
E = r. 

25. An ice cream cone is to hold one-eighth of a pint. The 
slant height is I, and half the angle at the vertex is x. Find 
the value of x that will make the cost of manufacture of the 
cone a minimum. (Ans. x = 35°.27.) 

8. Tangents in Polar Coordinates. Let 

be the equation of a curve in polar coordinates. We wish to 
find the direction of its tangent. The direction will be known 
if we can determine the angle tj/ between the radius vector pro- 
duced and the tangent. Let P, with the coordinates (tq, 6^, be 

an arbitrary point 
iT of the curve and 

P':(ro-FAr, flo+Afl) 
a neighboring point. 
Draw the chord 
PP and denote 
the Z OP'P by f . 
Then obviously 

lim^'= ^Q. 

To determine ^qj 
Pjq 45 drop a perpendic- 

ular PM from P on 
the radius vector OP' and draw an arc PN of a circle with 
as centre. The right triangle MP'P is a triangle of refer- 




TRIGONOMETRIC FUNCTIONS 129 

ence for the angle ij/' and 

^ MP 
Hence 

P'M 

(1) cot ^0 = li™- cot ij/' = lim • 

In the latter ratio we can replace P'M and MP by more 
convenient infinitesimals ; cf. Chap. IV, § 2. We observe that 

MP = Tq sin A6 ; hence lim = lim ; — = 1 ; 

i.e. MP and r^AB are equivalent infinitesimals. 

Furthermore, P'Jf and P'iV=Ar are also equivalent infini- 
tesimals. For _,,^ „,,_ , „,^ 
P'M=P'N+ NM 

and -?df =?■(, — ro cos Afl. 



1 — cos Afl 



Hence 



Now, by § 3, 



NM 


■'o- 


Ad 




Ar 




Ar 
AS 




limi^ 


- cos 


Afl 


0. 



A61 



On the other hand, ,• Ar _ j-. 

and this quantity is not, in general, 0. Hence 

lim^=0. 
A8^ Ar 

Returning to equation (1) we can now write the last limit in 

the form: p,^ ^ ^ 

hm = lim = — Dgr; 

p'=p MP Ae^roAe r^ 

or, dropping subscripts, 

(2) coti/r = i2),r. 



130 



CALCULUS 



In terms of differentials, this result can be written in either 
of the two forms : 



(3) 



cot^= 



dr 



tani/f = 



rdO 



rde' ' ^ dr 

Examvple. Consider the parabola in polar form 




TO 



1 — cos <^ 
To determine i/». Here, 



Hence 

C0tl/f = 



*"" (1-C0S^)2" 



m sin <^ d<f> 1 — cos <^ 
(1 — COS <^)2 md<^ 

_ sin<^ 

1 — COS ^ 

In particular, at the extremity 
of the latus rectum, we have : 

and thus we obtain anew the result that the tangent there 
makes an angle of 45° with the axis of the parabola. 
Again, at the vertex, 

COtU=„=0, ,/r = |, 

and the tangent there is verified as perpendicular to the axis. 
From the above equation, 

cot^ = ?^5^-, 

^ 1 — cos ^ 

a simple relation between ^ and <f> can be deduced. Since 
sin^ 



2 sin* cos* 

2 2,,^ 

^ — J — ; — =cot5, 

l-cos</, ,„.„,,^ 2 



2sin2 2 
2 



TRIGONOMETRIC FUNCTIONS 131 

it follows that , , .A 

eot^ = — cot^- 

But, for any angle, x, 

cot (w — x)= — cot X. 

Setting a; = ^ in the above equation, we have : 

cot (ir — i/f) = cot ^ • 

Hence * , <4 

or, fke supplement of ^ is equal to *. Thus we have a new 

proof of the familiar property of the parabola, that the tangent 
at any point P of the curve bisects the angle between the focal 
radius, OP, and a parallel to the axis, drawn through P. 

EXERCISES 

1. Plot the spiral, r=6, 

and show that the angle at which it crosses the prime direction 
whene = 27r is 80° 67'. 

2. Plot the spiral, _ 1 

Show that it has an asymptote parallel to the prime vector. 

Suggestion. Consider the distance of a point P of the curve 
from the prime direction, and find the limit of this distance 
when 6 approaches 0. 

Determine the angle at which the radius vector correspond- 
ing to 6 = 17/2 meets this curve. 

3. Plot the cardioid, 

r = a (1 — cos </>), 

* The trigonometric equation admits a second solution, namely, 
(^ _ ^) + T = <p/2. If, however, we agree to take and ^ so that 
■^ < 2 TT and '^ ^ < tt, this second solution is ruled out. 



132 CALCULUS 

and show that , , sin d> 

cot d/ = — — • 

1 — cos <l> 

At what angle is the curve cut by a line through the cusp 
perpendicular to the axis ? 

4. Prove that, for the cardioid, 

^ 2 

5. Show that the tangent to the cardioid is parallel to the 
axis of the curve when <^ = |^7r. 

6. At what points of the cardioid is the tangent perpen- 
dicular to the axis of the curve ? 

7. Determine the rectangle which circumscribes the cardioid 
and has two of its sides parallel to the axis of the curve. 

8. Show that, for the lemniscate, 

?-2 = a2cos2e, 
the angle i^ is given by the equation : 

cot i/' = — tan 26. 
Hence, show that , tt , o /i 

'I' = 2 

9. At what points of the lemniscate is the tangent parallel 
to the axis * of the curve ? 

Ans. At the point for which = 7r/6, and the points which 
correspond to it by symmetry. 

10. The points of the curve 

at which the tangent, is parallel to the prime vector, are evi- 
dently those for which 

•' y = r sm </), 

* The axis of any curve is a line of symmetry. The lemniscate has 
two such lines. The axis referred to in the text is that one of these lines 
which passes through the vertices of the curve. 



TRIGONOMETRIC FUNCTIONS * 133 

considered as a function of <^ through the mediation of the 
equation of the curve, has a maximum, a minimum, or a certain 
point of inflection. For these points, then, 

^ = r cos d + sin d. — = 0. 

Show that this condition is equivalent to the one used above 
in the special cases considered, namely : 

}fl -\- <^== IT. 

11. Plot the curve, r = acos2^, 
taking a= 5 cm. Show that for this curve 

cot./' = — 2 tan 26. 

12. At what points of the curve of question 11 is the tan- 
gent parallel to the axis ? 

Ans. For one of the points, tan 6 = 

V5 

13. Plot the curve, /• = a cos 3fl, 
taking o = 5 cm. Show that 

eot.^ = — 3tan3e. 

14. At what points of the curve of question 13 is the tan- 
gent parallel to the axis of the lobe ? 

Ans. For one of these points, tan =-v/l -| -• 

15. The equation _ m 

1 4- sin (/> 

represents a parabola referred to its focus as pole. Give a 
direct proof that the tangent to this curve at any point bisects 
the angle formed by the focal radius drawn to this point and 
a parallel to the axis through the point. 

16. Show that the tangent to the hyperbola 

m 
r = = 

1 — V3 cos <^ 



134 ' CALCULUS 

at the extremity of the latus rectum makes an angle of 60° 
with the transverse axis. 

17. Prove that the -tangent to the ellipse 

r=^ , 

V3 — cos ij) 

at the extremity of the latus rectum makes an angle of 30° 
•with the major axis. 

9. Differential of Arc. Let 

(1) y =/{<>') 

be the equation of a given curve. Let P, with the coordinates 
(x, y),he a variable point, and A a fixed point of the curve. 
Denote the length of the arc AF by s. Then s is a function 
of X ; for, when x is given, we know P and thus s. 

It is possible to determine the derivative of s, D^, as fol- 
lows. By the Pythagorean Theorem we have (Chap, IV, Fig. 33), 

PP^ = Aa!2 + ^y^'. 
Let A* approach as its limit. Then 

lim f^7= 1 + 1™ (¥)^= 1 +(Ojyy. 

Since by § 3 the chord PP' and the arc PP'= As are 
equivalent infinitesimals, it follows from the Fundame ntal 
Theorem of Chap. IV, § 2 that, in the above equation, PP' 
can be replaced by As. Hence 



limf^Y=limf^Y = (i).,)S 



and consequently 

(2) (D^sy=i+ip^yy. 



TRIGONOMETRIC FUNCTIONS 135 

On replacing the derivatiTes in (2) by their values in terms 
of differentials, we have 



\aaij \dxj 



or 

(3) ds^ = dx^ + dy^. 

This formula is easily interpreted geometrically by means 
of the triangle PMq, Fig. 33. Since 

PM= dx and MQ = dy, 

it follows from the Pythagorean Theorem that 

(4) PQ = ds. 

It is obvious geometrically that ds and As differ from each 
other by an iniiiiitesimal of higher order ; i.e. that they are 
equivalent infinitesimals.* 

Formulas for sin t, cos t. From the triangle PMQ we can 
write down two further formulas : 

/e\ • dy dx 

(5) sm T = -^ , cos T = — 

ds ds 

These formulas presuppose a suitable choice of t. As s in- 
creases, the point P describes the curve in a definite sense. 
Let this be chosen as the positive sense of the tangent line at 
P. Then t shall be the angle between the positive axis of x 
and this line. If t were taken as the. angle which the op- 
positely directed tangent makes with the positive axis of x, 
the — sign must be written before each right-hand side in (6). 

The formulas (6) suggest that x and y can be taken as func- 
tions of s : . , . ... 
X = ^(s), y = i/f(s). 

* In case the coordinates x and y are expressed as functions of a third 
variahle t, dx will not in general be equal to Aa;, but wiU differ from it by 
an infinitesimal of higher order. The triangle PMQ will then be replaced 
by a similar triangle PMiQi, in which Mi lies on the line PM, its distance 
from M being an infinitesimal of higher order. 



136 CALCULUS 

This is, of course, always possible, since, when s is given, P, 
and hence also x and y, are determined. 
Since 



(6) ds = ± \'dx' + dy , 
we have from (5) 

(7) sinT = +- — -^^ ■ cosr = ± ^ 



-s/da? + dy^ Vdx^ + dy'' 

no matter what choices of s and t are made.* Furthermore, 

dy 
(8) 3inT = + — '^^ cosT = ± ^ 

Which sign is to be used in (8) depends on which of the two 
possible determinations has been chosen for t. Thus t in a . 
given cas^ might be 30° or 30° + 180° = 210°. If the first 
choice were made, t = 30°, then sin. t, cos t, and dy/dx = tan t 
would all be positive quantities, and hence the upper signs 
must be taken. But if the other choice, t = 210°, is made, then 
sin T and cos t are negative, and the lower signs hold. 

Example. Consider the parabola 

y = x\ 

Let P be a point of the curve which lies in the first quadrant. 

Since 

tanT = ^ = 2a; 
dx 

is here positive, t may be taken as an angle of the first quad- 
rant. In that case, formulas (8) give 

2a! 1 

smT = — . cost = 



Vl + 4a;2 VH-4a;2' 

* The signs in (6) and (7) are not necessarily tlie same ; also in (7) and (8). 



TRIGONOMETRIC FUNCTIONS 137 

If P is a point of the curve which lies ia the second quad- 
rant, tan T is negative, and t is an angle of the second or fourth 
quadrant. If we choose to take t as an angle of the second 
quadrant, formulas (8) become 

■ 2x 1 

sinT = . eosT = - 



Vl + 4aj2 Vl+4a;2 

We may, however, equally well take t as an angle of the fourth 
quadrant. Then 

sinT= — . cost = - 



Vl+4a;' Vl + 4a!2 

In each ease, one of the numbers, sin t and cos t, is positive, 
the other, negative. 

Polar Coordinates. Similar considerations in the case of 
the curve „.„. 

lead to the following formulas ; cf . Fig. 46 : 
PP'^ = P'3P + MP^. 



limf^Y=limr^%limf^' 



Hence 



Now, the chord PP and the arc PP' = As are equivalent 
infinitesimals. Moreover, PM and Ar are equivalent ; and 
MP and r^B are equivalent. Hence 

« 

Dropping the subscript and writing the derivatives in terms 

of differentials we have, then : 

or 

(10) ds2 = dr-2 + r^dB\ 



138 CALCULUS 

Furthermore, 
(11) .i«t=f, c<»*.|, 

the tangent PT being drawn in the direction of the increasiag 
s, and ^ being taken as the angle from the radius vector pro- 
duced to the positive tangent. 

10. Rates and Velocities. The principles of velocities and 
rates were treated in Chapter III, § 8. We are now in a 
position to deal with a wider range of problems. 
' We note the following formulas. Let a point P describe 
the curve „, ^/„v 

Let s denote the length of the arc, measured from an arbitrary 
point in an arbitrary sense, and let t be the angle from the 
positive direction of the axis of x to the tangent at P drawn 
m the direction of the increasing arc. Then the components 
of the velocity (v = ds/dt) of P along the axes are, respectively: 

-H, . da; dy . 

(1 ) — = v cos T, -s-= V sm T. 
^ ^ dt ' dt 

Let a point P describe the curve 

(2) r = ¥{&). 

Let s denote the length of the arc, measured from an arbitrary 
point ia an arbitrary sense ; and let \ji be the angle from the 
radius vector, produced beyond P, to the tangent at P drawn 
iu the direction of the increasing arc. Then the components 
of the velocity {v = ds/dt) of P along the radius vector pro- 
duced and perpendicular to the same (the sense of the increas- 
ing 6 biing taken as positive for the latter) are respectively : 

^ ^ dt ^ dt ^ 

Example 1. A railroad train is running at the rate of 30 
miles an hour along a curve in the form of a parabola : 

y2 = 1000 a;, 




TRIGONOMETRIC FUNCTIONS 139 

the axis of the parabola being east and west, and the foot being 
taken as the unit of length. The sun is just lising in the east. 
Find how fast the shadow of the locomotive is moving along 
the wall of the station, which is north and south, when the 
distance of the shadow from the axis of the parabola is 300 ft. 
The first thing to do is to draw a suitable figure, introduce 
suitable variables, and set down all the data not already put 
into evidence by the figure. Thus in the 
present case we have, in addition to the ac- 
companying figure, the further data : (a) the 
velocity of the train ;' this must be expressed 
in feet per second, since we wish to retain 
the foot as the unit of length for the equa^ 
tion of the curve. Now, 30 miles an hour 
is equivalent to 44 feet a second. On the other hand, another 
expression for the velocity is ds/dt. Hence we have, on 
equating these two values, 

^=44 
dt' 

(b) We must set down explicitly at this "point the equation 
of the curve, „ . ^^^ 

To sum up, then, we first draw the figure and then write 
down' the supplementary data : 

Given a) -- = 44, 

dt 

and' b) y^ = 1000x. 

The second thing to do is to make clear what the problem 
is. In the present case it can be epitomized as follows : 

Tofind (^] ■ 

We are noj7 ready to consider what methods are at our dis- 
posal for solving the problem. We observe that ds occurs in 



140 CALCULUS 

the data. Obviously, then, we must make use of the one gen- 
eral theorem we know which gives an expression for ds when 
the equation of the curve comes to us in Cartesian coordinates, 
— namely, the theorem : 

ds' = da? + dy\ 

Since dx occurs neither in the data nor in th§ conclusion, 
we wish to eliminate it. This can be done by means of the 
equation of the path 6). DifiEerentiating 6) we have : 



Hence 



dy 


= 1000 da;. 


dx 


_ydy , 

500 



Consequently ds^ = "Jf + ( 



500" 



and 



'=<-i^+^^y- 



5002 

The next step is obvious ; divide through by dt : 



ds^ I y I idy 
dt ^ 5002 "^ at 

In this last equation, replace ds/dt by its known value from 
a), and we now have an equation for determining dy/dt : 

dy 44 

\ 5002 ^ 

Finally, bring into action the particular value of y with 
which alone the proposed equation is concerned, namely, 

y-'"'- /^N _ 44 ^ 44 _3,,3^ 
\dtjy,3a> V.62 + 1 VT36 

or, the rate at which the shadow is moving along the wall of 
the station is 37.73 ft. a second. 



TRIGONOMETRIC FUNCTIONS 141 

Angular Velocity. By the angular velocity, u>, with which a ' 
line is turning in a given plane is meant the rate at which the 
angle, ^, made by the rotating line with a fixed line, is in- 
creasing : 

do 

dt 

Example 2. A point is describing the eardioid 
r = a (1 — cos 6) 

at the rate of c ft. a second. Find the rate at which the 
radius vector drawn to the point is turning when 6 = ir/2. 
The formulation of this problem is as follows : 

r — a(l — cos ff).* 



Given 


a) 


and 


V) 


To find 




Since, from § 9 (10), 



and from 6), ' 

it follows that ' 

ds2 = a2 sin2 ed&^ + a\l — cos fff d&^ 
— a'^de^[sm^6 + 1 — 2 cos ^ + cos^S] 

=2 d'dff^ • (1 - cos-(9) = 4 a'' sin^ ^ dff^. 
Hence, s being measured from the cusp, 

ds = 2asin-dfl, 
2 ' 

J ds rt ■ BdB 

and — = 2 a sm 

dt 2 d« 

* The student should make a free-hand drawing of the curve. 



142 


CALCULUS 


Consequently, by 


the aid of a) 




ae c 




^' 2asinf 
2 


and thus, finally 


fde\ _ c 



»V2 



EXERCISES 

1. A point describes a circle of radius 200 ft. at the rate 
of 20 ft. a second. How fast is its projection on a fixed 
diameter travelling when the distance of the point from the 
diameter is 100 ft. ? Ans. 10 ft. a second. 

2. A flywheel 16 ft. in diameter is making 3 revolutions a 
second. The sun casts horizontal rays which lie ia or are 
parallel to the plane of the flywheel. A small protuberance 
on the rim of the wheel throws a shadow on a vertical wall. 
How fast is the shadow moving when it is 4 ft. above the 
level of the axle ? 

3. A revolving light sends out a bundle of rays that are 
approximately parallel, its distance from the shore, which is 
a straight beach, being half a mile, and it makes one revolu- 
tion in a minute. Find how fast the light is travelling along 
the beach when at the distance of a quarter of a mile from the 
nearest point of the, beach. 

4. A point moves along the curve r = 1/6 at the rate of 
6 ft. a second. How fast is the radius vector turning when 
6 = 2w? 

5. In the example of the ladder. Chap. Ill, § 8, Ex. 5, find 
how fast the ladder is rotating at the instant in question. 

6. At what rate is the direction of the second ship from the 
first changing at the instant in question, in Ex. 2 of Chap. Ill, 
§8? 



TRIGONOMETRIC FUNCTIONS 



143 




7. How fast is the direction of the man from the lamp- 
post changing in Ex. 12 of Chap. Ill, § 8 ? 

8. The sun is just setting aa a baseball is thrown vertically 
upward so that its shadow mounts to the highest point of the 
dome of an observatory. The dome is 50 ft. in diameter. 
Find how fast the shadow of the ball is moving along the 
dome one second after it begins to fall,' and also how fast it is 
moving just after it begins to fall. 

9. Let AB, Fig. 48, represent the rod that connects the 
piston of a stationary engine with the fly-wheel. If u denotes 
the velocity of A in its rectilinear path, 
and V that of B in its circular path, / «/V^~-— i 
show that 

M =:(sin 6 + cos 6 tan ^)?;. 

FiQ. 48 

10. rind the velocity of the piston of 

a locomotive when the speed of the axle of the drivers is given. 

11. A drawbridge 30 ft. 
long is being slowly raised 
by chains passing over a wind- 
lass and being drawn in at 
the rate of 8 ft. a minute. A 
distant electric light sends 
out horizontal rays and the 
bridge thus casts a shadow 
on a vertical wall, consisting of the other half of the bridge, 
which has been already raised. Find how fast the shadow 
is creeping up the wall when half the chain has been 
drawn in. i 

12. A man walks across the floor of a semicircular rotunda 
100 ft. in diameter, his speed being 4 ft. a second, and his 
path the radius perpendicular to the diameter joining the 
extremities of the semicircle. There is a light at one of the 
latter points. Find how fast the man's shadow is moving along 
the wall of the rotunda when he is halfway across. 




Fig. 49 



144 CALCULUS 

13. A man in a train that is running at full speed looks out 
of the window in a direction perpendicular to the track. If 
he fixes his attention successively for short intervals of time 
on objects at different distances from the train, show that the 
rate at which he has to turn his eyes to follow a given ohject 
is inversely proportional to its distance from him. 

14. Water is flowing out of a vessel of the form of an in- 
verted cone, whose semi-vertical angle is 30°, at the rate of a 
quart in 2 minutes, the opening being at the vertex. How 
fast is the level of the water falling when there are 4 qt. of 
water still in ? 

15. Suppose that the locomotive of the first of the Examples 
worked in the text is approaching the station at night at the 
rate of 20 miles an hour, its headlight sending out a bundle 
of parallel rays. How fast will the spot of light be moving 
along the wall of the station when the distance of the head- 
light from the vertex A of the parabola, measured in a straight 
line, is 500 ft. ? 

Assume that the wall is perpendicular to the axis of the 
parabola and distant 75 ft. from the vertex. 

16. In the preceding question, how fast will the bundle of 
rays be rotating ? 

17. A point describes a circle with constant velocity. Show 
that the velocity with which its projection moves along a given 
diameter is proportional to the distance of the point from this 
diameter. 

18. A point P describes the arc of the ellipse 

which lies in the first quadrant, at the rate of 12 ft. a second. 
The tangent at F cuts off a right triangle from the first quad- 
rant. How fast is the area of this triangle changing when P 
passes through the extremity of the latus rectum ? Is the area 
increasing or decreasing ? 



TRIGONOMETRIC FUNCTIONS 145 

19. A point P describes the oardioid 

r = 5 (1 — cos ff) 

at the rate of 12 cm. a second. The tangent at P cuts the 
axis of the curve in Q. How fast is Q moving when 6 = 7r/2 ? 

20. The sun is just setting in the west as a horse is running 
around an elliptical track at the rate of m miles an hour. The 
axis of the ellipse lies in the meridian. Find the rate at 
which the horse's shadow moves on a fence beyond the track 
and parallel to the axis. 



CHAPTER VI 



LOGARITHMS AND EXPONENTIALS 

1. Logarithms. The logarithms with which the student is 
familiar are those which are ordinarily used for computation. 
The base is 10, and the definition of logioic is as foUows : 

y = logio X if 10" = X. 

These are called denary, or Brigg^s, or common logarithms. 

More generally, any positive number, a, except unity, caai 
be taken as the base, the definition of log, x then being : 

(1) 3/ = log, a; if a«=x. 

y = log I 




Fig. 50 



The accompanying figure represents in character the graph 
of the function log, x for any a > 1. It is drawn to scale for 

146 



LOGARITHMS AND EXPONENTIALS 147 

a = 2.71828. The reason for this choice of a will appear 
shortly. 

From the definition it follows at once that 

(2) log.l = 0, log„a = L 

Only positive numbers have logarithms. For, a" is always 
positive. Hence, if x be given a negative value (or the value 
0), the second equation under (1) above cannot be satisfied by 
any value of y. 

The two leading properties of logarithms are expressed by 
the equations : * 

(I) logP+logQ = log(PQ) 

(II) logP" = nlogP. 

Here, P and Q are any two positive numbers whatever, and n 
is any number, positive, negative, or zero. The base, a, is 
arbitrary. Thus 

log 10 = log 2 + log 5 

and log VT = log 7^ = J log 7. 

From equation (I) it follows that 

(3) log^ = -logQ 
and 

(4) log 1= log P- log Q. 



For, if we set P = 1/Q in (I), we have 
But, by (2), 



log 1 = log - + log Q. 



log 1 = 0. 

* The student should recall the proofs of these theorems, which he 
learned in the earlier study of logarithms, and make sure that he can 
reproduce them. Proofs of the theorems are given in the author's 
DWerential and Integral Calculus, p. 76. 



148 CALCULUS 

Hence i 1 i /-» j 

log- = -logQ, q.e.d. 

Again, write (1) in the form 

logiPQ') = logP+logQ', 

and now set Q' = 1/Q. Then 

P i 

log- = logP+log-. 

But logi = -logQ. 

Hence 

p 
log-=logP-logQ, q.e.d. 

For example, 

log (a + b)- log a = log fl + ^\ 

as we see by setting, in equation (4), 

P=a + b, Q = a. 

As a further example of the application of equation (II) we 
may cite the following : 

^°g ^t + ^) = log Ka + ¥}- 
ft 

Eor, if P=a + 6 and »=-, the left-hand side of this equa^ 

h 

tion has the value n log P. 

A Further Property of Logarithms. When it is desired to 
express a logarithm given to a certain base, a, in terms of 
logarithms .taken to a second base, b, the following relation is 
needed : 

(III) log.a; = j5Si^. 

The proof of (III) is as follows. Let 

y = log» X, a' = X. 



LOGARITHMS AND EXPONENTIALS 149 

Take the logarithm of each side of this equation to the base b : 

(5) logj a» = logt X. 

But the left-hand side can be transformed by (11)^ if in (II) 
we take 6 as the base, thus having 

logjP»=nlogiP. 
Here, let 

P=a, n = y. 

Then logj a<'=y log^ a, 

and (5) now becomes : 

y logj a = logi X. 

y = ^^, or log„., = |2i^, q.e.d. 

logs a logs* 

Example. Let 6 = 10 and let a = 2.718. To iind log„ 2. 
From (III), 

log 2 = _i2gio2_ _ .3010 ^ gg32 
^° logio 2.718 -4343 

' Two Identities. Just 'as, for example, 

-\/q^ = x and (Va:)' = «, 

no matter what value x may have, so we can state two identi- 
ties for logarithms and exponentials. In the second equation 
(1), replace y by its value from the first equation. Thus the 
equation 

(6) rf»So"- = a! 

is seen to hold for all positive values of x. 

Secondly, replace x in the first equation (1) by its value 
from the second equation : 

y = log„ a". 

We can equally well write x instead of y, understanding 
now by x any number whatever, and we have, then, the 



150 CALCULUS 

identity 

(J) log. a' = X. 

This equation holds for all values of x, positive, negative, or 
zero. 

EXERCISES 

1. Show that , -.„. -._„ 

logio .8950 = — .0482. 

2. rind logio .09420. -4ns. -1.0259. 

3. Compute 2.718-6M2. Ans. 1.758. 

4. Compute 2.718--«'^°. Ans. 0.4186. 

5. Compute w''. 6. Compute V2*'^_ 

7. Show that 

log tan 6 = log sin 6 — log cos ^, <6 <-■ 

8. Show that 

log sin e + log cos fl = log 55:^, o<e<l. 

9. Show that 

j^gl-|ose_21ogsin|, 0<fl<2,r. 

10. If (a;, y) are the Cartesian coordinates of a point distinct 

from the origin, and (r, ff) the polar coordinates of the same 

point, show that , , , , „ « 

^ ' log r = 1^ log (as' + 2^'). 

11. Prove that 

log (a^ - 62) = log (a + 6) + log (a - 6), 
provided a + 6 and a — b are both positive quantities. 

12. Simplify the expression 

log (1 + ;»•)- log (l+a!2). 

13. Show that 

V(e^ — e-')2 + 4 = e' + e"*, 

where e has the value 2.7182. 



LOGARITHMS AND EXPONENTIALS 151 

14. Simplify the expression 



15. Show that 

ilog(l + = log(H-0'. 

2. Differentiation of Logarithms. In order to differentiate 
the function , 

it is necessary to go back to the definition of a derivative, 
Chap. II, § 1, and carry through the process step by step. 

Give to X an arbitrary positive value, Xq, and compute the 
corresponding value, 2/0, of the function : 

(1) VO = loga Xg. 

Next, give to x an increment Ax (subject merely to the restric- 
tion that Xq + Ax is positive and Ax # 0) and compute the new 
value, 2/0 + Ay, of the function : 

(2) yo + Ay= log„ (xa + Ax). 

From (1) and (2) it follows that 

Ay ^ log, (xq + Ax) — log„ x„ 
Ax Ax 

It is at this point that the specific properties of the loga- 
rithmic function come into play for the purpose of transform- 
ing the last expression. By § 1, (4), 



log„ {xo + Ax) - log„ xo = log„ fl + ^\ 



and hence 



We next replace the variable Ax by a new variable t as 
follows : 



152 CALCULUS 

t=— or Aa; = x^. 

Xf, 

Thus (3) takes on the form 

From (II), § 1, the bracket is seen to have the value 
log. (1 + OS 



and hence 



^ = ilog„(l + 0'. 



(4) 

Ax Xg 

As Aa; approaches as its limit, t also approaches 0, and so 
(6) limf2^ = llimlog,(l + 0^- 

i_ 
Now, the variable (1 + ' approaches a limit when * ap- 
proaches 0, and this limit is the number which is represented 
in mathematics by the letter e; cf . § 3. Its value to five 
places of decimals is 

e = 2.71828 .•• ; 

cf . § 3. Moreover, log a; is a continuous function of x, as is 
shown in a detailed study of this function.* Hence 

lim log„ (1 + 0'"= log„ { lim (1 + *)4 = log« «■ 

On substituting this value in the right-hand side of (5) we 
have: 

Xq 

* Such a treatment is too advanced to be pursued with profit at this 
stage. Cf. the author's Differential and Integral Calculus, Appendix, 
p. 417. 



LOGARITHMS AND EXPONENTIALS 153 

or, on dropping the subscript : 

(6) D,log„a; = ?:5g^. 

X 

Thus if the usual base, a = 10, be taken, the formula 
becomes : 

f7\ n 1 .4343... 

(7) Z)Jogioa!= 

X 

Discussion of the Result. We have met a similar situation 
before, in the differentiation of the sine. There, if angles be 
measured in degrees, the fundamental formula reads : 

J5, sin X = -^ cos x. 
180 

In order to get rid of this inconvenient mxdtiplier, we 
changed the unit of angle from the degree to the radian, and 
then the formula became : 

Dj sin X = cos x. 

In the present case, it is possible to do a similar thiag. The 
base, a, is wholly in our control, to choose as we like. Now, 
for any base, the logarithm of the base is unity, § 1, (2) : 

log„ a = 1. 

If, then, we choose as our base the number e : 

a = e = 2.71828 - 

the multiplier becomes 

(8) log„ e = log. e = 1. 

For this reason, e is taken as the base of the logarithms 
used in the Calculus.* These are called natural logarithms. 
They are also called hyperbolic, or Naperian logarithms, — the 
latter name after Napier, the inventor of logarithms. But 

* The notation e for this number is due to Euler, 1728. 



154 CALCULUS 

Napier * was the very man who introduced denary logarithms 
into mathematics, and so the use of his name in connection 
with natural logarithms is misleading. 

Since natural "logarithms are always meant in the formulas 
of the calculus, unless the contrary is explicitly stated, it is 
customary to drop the index e from the notation log, a; and 
to write 

(9) y = log X, if e" = x. 

The identities (6) and (7) of § 1 now take on the form : 

(10) 6'°^" = X, 

(11) log e' = X. 

The formula of differentiation becomes : 



(12) 


DJogx = -- 

X 


In differential form it reads : 


(13) 


4-logx=^, 
dx X 


(14) 


dlo8x = ^. 

X 


Example. 


Differentiate the function 




u = log sia X. 


Let 


y = sin X. 


Then 


u = logy, 


t 


Z«. = rf;ioB-i/=^. d 



dy = cos xdx, 

y 

and 

, cosxdx . , 

du = — : = cot X ax. 

siniB 

* Napier was a Scotchman, and Ms discovery was published in 1614. 



LOGARITHMS AND EXPONENTIALS 155 

Hence d log sin x = cot x dx, 

or — log sin x = cot x. 

dx 



EXERCISES 

Differentiate the following functions. 

1. M = log cos X. -r- = — tan x. 

dx 

2. M = log tan x. — = cot x + tan x. 

die 

du -2 



3. M = log cot «. , — . r> 

aa; sinzoj 

4. M = log sec K. 5. M = log CSC a;. 

„ , 0! du 1 , 1 

6. u = log:j — =- + 



7. M = log 



1 — a; da!a!l — K 

a + a; du _ 2a 

a — a; da; a^ _ a;2 

(Zm a; 



8. M = log Va^ + xK 

_ 1 /< \ dw . a; 

9. M = log (1 — cos a;). — = cot - 

da; 2 

10. u = log (1 + cos a;). T'~~ ^^^ n 



dx a? + x^ 

du .x 

— = cot — 
da; 2 

du , 

— = — tan 
da; 2 



3. The Limit lim (1 + t) '- Since this limit is fundamental 

in the differentiation of the logarithm, a detailed discussion 
of it is essential to completeness. Let us set 

(1) s=(l + 0^ 

and compute the value of s for values of t near 0. Suppose 
t = .1. Then 

s= (1.1)10, 



156 CALCULUS 

and this number is found by the usual processes with loga- 
rithms to be 2.69. 

Further pairs of corresponding values {t, s) are found in a 
similar manner. In particular, the student can verify the 
correctness of the following table of values : * 



t 


-0.1 


-.01 


-.001 . 


. + .001 


+ .01 


+ 0.1 


s 


2.87 


2.73 


2.72 . 


. 2.72 


2.70 


2.59 



The foregoing table indicates strongly that, when t ap- 
proaches the limit from either side, the variable s is 
approaching a limit whose value, to three significant figures, is 
2.72. This is ia fact the case.f The exact value of the limit 
is denoted by the letter e : 

I 
(2) lim (1 + «)' = e = 2.71828 -. 

4. The Compound Interest Law. The limit (2) of § 3 pre- 
sents itself in a variety of problems, typical for which is that 
of finding how much interest a given sum of money would 
bear if the interest were compounded continuously, so that 
there is no loss whatever. For example, $ 1000, put at in- 
terest at 6%, amounts in a year to $1060, if the interest is 
not compounded at all. If it is compounded every sis months, 

we have , f^/., 

$1000(1 -I- 4^ j 

as the amount at the end of the first six months, and this 
must be multiplied by [l4-^| to yield the amount at the 
end of the second sis months, the final amount thus being 

$iooo/'i + --^^ 

* To compute the middle entries in this table a six-place table of 
logarithms is needed. 

t For a rigorous proof cf . the author's Differential and Integral CaJr 
cuius, p. 79. 



LOGARITHMS AND EXPONENTIALS 157 

It is readily seen that if the interest is compounded n times 
in a year, the principal and interest at the end of the year will 
amount to , „„. „ 

1000(^1 + :^) 

dollars, and we wish to find the limit of this expression when 
«. = 00. To' do so, write it in the form : 



1000 



n 



and set t=' — . The bracket thus becomes 
n 

and its limit is e. Hence the desired result is 
1000e»« = 1061.84.* 



EXERCISE 

If $ 1000 is put at interest at 4 % , compare the amounts of 
principal and interest at the end of 10 years, (a) when the 
interest is compounded semiannually, and (6) when it is com- 
pounded continuously. Ans. A diEEerence of $ 6.88. 

5. Differentiation of e'. Before beginning this paragraph 
the student wUl turn to Chap. VIII and study carefully § 1. 
Since 

(1) y = ^ and X = log y 

are equivalent equations, the former function can be differen- 
tiated by taking the differential of each side of the latter 

equation: ^ 

ax = diogy = -^- 

y 

* The actual computation here is expeditiously done by means of 
series ; see the chapter on Taylor's Theorem. 



158 
Hence 
or 
(2) 

(3) 



The function 



CALCULUS 


dy _ 
dx 


y, 


d^ 
dx 


«*, 


de = 


e'cte. 



y 


= e' . 




.1 




X 










Fig. 51 



' = a' 



(4) 

could be differentiated in a similar manner. It is, however, 
simpler to take the logarithm of each side of (4) and then dif- 
ferentiate the new equation : 

logy = log a* = a; log a, 
d log w = -=^ = da; log o. 

y 



(5) 



da' = al'logadx. 



Differentiation of a!". It is now possible to complete the dif- 
ferentiation of this function for the case that n is irrational. 



LOGARITHMS AND EXPONENl 


Since by § 2, 


(10), 


X = e'-s', 


it follows that 




x' = e"^"^', 


and hence 




= e''^''^'d(nlogx) 
-eni^e^ndx 

X 


Thus finally, 




X 


(6) 




dx'" = «a!"~idx, 



159 



no matter what value n may have, provided merely that n is 
a constant. 

Differentiation of /(a!)'"'''. Let it be required, for example, 
to differentiate the function 

2/ = af. 

Here, both base and exponent are variable. Begin by tak- 
ing the logarithm of each side of the equation : 

log y = log X' = X log X. 

Hence ,, ,, , 

a log y = d{fe log x), 

°^ ^ = {l + \ogx)dx, 

y 

and so, finally, 

das' = a?(l + log x)dx. 



dy = y{l + log X) dx 
or 



The general case, „ _.y(^a;')*' 

can be treated in a similar manner. 



160 



CALCULUS 



6. Graph of the Function «". Por positive values of n the 
curves 

y = x' 




Fis. 52 



lie as indicated in the figure. When n = 1, we have the ray 
from the origin, which bisects the angle between the positive 
axes of X and y. 



LOGARITHMS AND EXPONENTIALS 161 

When n > 1, the curve is always concave upward ; when 
w < 1, it is concave downward. 

All the curves start at the origia and pass through the 
point (1, 1). 

Por values of a; > 1, the larger n, the higher the curve lies. 
For values of a; < 1, the reverse is the ease. 

Let X have any fixed value greater than unity : a; = a;' > 1. 

Consider the ordiaate ,„ 

y = x'". 

As n increases, x'" increases continuously. This property is 
the basis of the property of logarithms included m the word 
continuous. 

For proofs of the foregoing statements of. the author's 
Differential and Integral Calculus, p. 27 and Appendix, p. 417. 

7. The Fornmlas of Differentiation to Date. The student 
will now bring his card of formulas up to date by supple- 
menting it so that it will read as follows : 

Genebal Formulas of Diffbrbsttiation 

I. dcu = c du. 

IL d(u + v) = du + dv. 

III. d (uv) =udv + v du. 

TV ,fu\_ vdu — udv 

\vj ifl 

Special Formulas of Differentiation 

1) dc = 0. 

2) da;'' = wa;"~ida;. 

3) d sin X = cos. x dx. 

4) d cos a; = — sin x dx. 

5) dtan x = sec^ x dx. 



162 CALCULUS 

6) dcota! = — csc^a; da;. 

7) dloga; = ^. 

X 

8) d^ = ^dx. 

9) da" = a" log a dx. 

To obtain facility in the use of the new results it is desirable 
that the student work a good number of simple exercises. 

Example 1. To differentiate the function 

u = e"'. 

Let y = ax. 

Then u = e", 

du = de" = e''dy = e"(adx). 
Hence 

-"'■ = ae' 



' de" 


= ae°*da; or — e' 
dx 


Example 2. 

show that * 


If 


u = Aco8 (nt + y), 



To do this, compute first — . The computation is readily 

az 

effected by taking the differential of each side of the given 

equation : 

du = Ad(X)s(nt+y) 

= A£—sia{nt + y)d(nt+ y)] 

= — An sin (nt + y) dt, 

* Such an equation as the following is called a differential egvation, 
and any function which, when substituted for u, satisfies the equation is 
called a solution. 



LOGARITHMS AND EXPONENTIALS 163 



— = — An sin {nt + y). 
Next, compute — . Since 



dH_d^du\ 

dP~di\dt) ~~dF 

dM_ 
dt' 



/ dt 
we take the differential of each, side of the equation for 

df — - J = — An d sin (nt + y) 

— — An [cos {nt + y)d{nt + y)] 
= — An^ cos (nt+y) dt. 
Hence, on dividing through by dt, we have : 

^ = -An^ COS (nt + y). 

If now we multiply the given value of u by n^ and add the 

dhi 
product to the value just obtained for — , the result is iden- 
tically 0, i.e. for all values of t : 

— + n^M = 0, q. e. d: 

dt^ ' ^ 

EXERCISES 

Differentiate the following functions. 

1. u = e-''. ~ = -2xe-". 

dx 

dit 

2. w = e='"^ — =e™»^cosa;. 

da; 

3. M = (e» + e--)2. ^ = 2(e2'-e-2'). 

dx 



164 CALCULUS 

4. M = 10'. — = (2.30259 OlO*. 

da; 

5. w = a!i<'10'. ^=»9 10-(10 + 2.30259a!). 

dx 

6. u = log (sec X + tan x). — = see x. 

dx 

7. M = a!2 1oga;. — = a;(l + 2 loga;). 

dx 

8. M = a' log (a — a;). 9. m = e"'log(2a; + 3). 

10. u = e""' COS (nt — y). 11. u = e~«*(^ cos ni-f-B sin n<). 

a; log a; , / , i\ dw loga; 

12. M= — i^:^ log(a; + l). — = . f^, • 

a; + l da; (a; + 1)2 

dw 1 



13. u = log(a;+Va;2_a2). ^ = 



14. u = log (as + Va^ + a 

15. u = log (e* + e^). 

17. M = log tan- • 

18. M = logtan^| + j\ 



da; Va;" — a? 
du 1 



19. „ = cotg-| 



20. u 



21. M = log VI + sin 0. 



23. u = log Vl — cos X. 
25. M=(10i+')=. 

27. «=r^Y- 





da; 


Va2 + a;'' 


16. 


M = 


sin a; + cos x 


e' 




du 
dx' 


= CSC X. 




du 
do'' 


= sec 6. 




du 


1 




dx 


1 — sin a; 




du 


1 




dx 


1 + sina; 


22 


. M = 


= log«^^. 

X 


24 


. U = 


=v^. 


26 


. Mi 


= </^. 


28, 


, U = 


= VIO'. 



LOGARITHMS AND EXPONENTIALS 165 

29. u = af"". ^ = a!'"'»'-i(sina; + a!cosa!loga;). 

30. M = (siiia;)'=°". ^= (suia!)™»--i(cos2a!- sitfajlogsina;). 

31. u = a;"'. 32. u = (cos a;)*'""^. 33. u — (tan x)". 

X 

34. •M = (loga;2)^ 35. m = (1 + a)«. 36. u = {x^f. 

37. If M = J. cos ni + JB sin nJ, show that 



38. If M = Oe-" cos (Vw^ — K^f + y), show that 

h 2 K h w% = 0. 

d«2 df 



CHAPTER VII 
APPLICATIONS 

1. The Problem of ITninerical Compntation. It often 
happens in practice that we wish to solve a numerical equa- 
tion in one unknown quantity, or a pair of simultaneous 
equations in two unknowns, to which the standard methods 
with which we are familiar do not apply ; for example, 

cos x = X, 
f 2 cot fl + 2 = cot <^, 
[ 2 cos + cos <f> = 2. 

Such equations usually come to us from physical problems, 
and the solution is required only to a limited degree of 
accuracy, — say, to two, three, or possibly four significant fig- 
ures. Any method, therefore, which yields an approximate 
solution correct to the prescribed degree of accuracy furnishes 
a solution of the problem. 

In particular, the problem of the determination of the 
error in the result due to errors in the observations comes 
under this head. 

2. Solution of Equations. Known Graphs. 
Example 1. Let it be required to solve the equation 

1) cos x = x. 

We can evidently replace this problem by the following: 
To find the abscissa of the point of intersection of the curves 

2) y = cos x, y = x. 

166 



APPLICATIONS 167 

The first of these curves we have plotted accurately to scale. 
The second is the right line through the origin, which bisects 
the angle between the positive coordinate axes. It is, there- 
fore, sufficient to lay down a ruler on the graph of the former 
curve, so that its edge lies along the right line in question, 
and observe where this line cuts the curve. The result lies 

between _ , 

x= .1 and X = .8, 

and may fairly be taken as a; = .76. It is understood, as 
usual iu approximate values, that the last figure tabulated does 
not claim complete accuracy ; but we are entitled to a some- 
what better result than would be given by the first figure 
alone. 

'Example 2. To solve the equation 

3) a!3-f-2a;-2 = 0. 

Suppose we have plotted the curve 

4) y = 0? 

accurately from a table of cubes. Then the problem can con- 
veniently be formulated as follows : 

To find the abscissa of the point of intersection of the curves 

5) y — s? and y — 2 — 2x. 
The details are left to the student. 

Example 3. To find the positive root of the equation 

6) e-i''4- 2.92 a; = 2.14. 

Here, we can connect up with the graph of the function e* 
by making a simple transformation. Let 

7) x'=: — ^x; x = — 2x'. 
The equation then becomes 

8) es'' - 5.84a;' = 2.14, 



168 CALCULUS 

and we seek to determine the abscissa of that point of inter- 
section of the curves (for simplicity, we drop the accent) 

9) ^ = 6^ and y = 5.84a; + 2.14 

which lies to the left of the origin. The second place of 
decimals in the coefllcients is not to be taken too seriously ; we 
make as accurate a drawing as the graph and a well-sharpened 
pencil permit. Having thus determined the negative a' from 
the graphs of 9), we find the desired positive x by substituting 
this value in equations 7). The execution of the details is 
left to the student. 

Example 4. Solve the equation 

e* = tan x, < a; < ~ / 

If one of the curves 

2/ = e* or y=. tan x 

were plotted on transparent paper, or celluloid, it could be 
laid down on the other with the axes coinciding and the inter- 
section read off. The same result can be attained by holding 
the actual grapTis up in front of a bright light. 

In cases as simple as this, however, free-hand graphs will 
often yield a good first approximation, and further approxima- 
tions can be secured by the numerical methods of the later 
paragraphs. 

EXERCISES* 

1. Solve the equation 

cos X = 2x. 

2. Find the root of the equation 

3 sin x = 2x 
which lies between and tt. 

* In solving these exercises only so great accuracy is expected as can 
be attained from well-drawn graphs of the standard cviTves. It will be 
shown in later paragraphs how the solutions can be improved analytically 
and carried to any desired degree of accuracy. 





APPLICATIONS 


169 


3. Solve: 


X + tan a; = 1, 


o<.<|. 


4. Solve: 


3 cos a; — 6a! = 6, 


-|<.<o. 



5. Find the root of the equation 

log x^! + 2 = a; 
which lies between and 1. 

6. Solve : sin 2a! = x. 

7. Find all the roots of the equation 

12a!' + 4a! + 3 = 0. 

8. The same for „ , „ ^ r. 

K)'ie — t>x — 1 = 0. 

9. The same for 

ar* — a; — 1 = 0. 

Solve the following equations : 

10. cos' e + .47 cos - 1.23 = 0, < 6 < 90°. 



11. sin X = Vl — x^. 12. a;2 + cos^ a; = 4. 

13. Show that the equation 

tan a; = a; 

has an infinite number of roots. These can be written in the 

form 

x„ = Wff + e„, 

where e„ is numerically small when n is numerically large. 

14. Find the largest value of P for which the equation 

cos X + Px = 1 
admits a solution in the interval < a; < ir. 

15. Find the point of the parabola 

2y = x^ 
which is nearest to the point (2, 0). 



170 CALCULUS 

16. rind the radius of the circle whose center is at (0, 2) 
and which is tangent to the parabola 

y^ = x. 

3. Interpolation. Consider the equation 

1) f{x)=0. 

Suppose a root has been located with some degree of accuracy. 
More precisely, suppose that 

/(rBi) and f{x^ 

are of opposite signs. If the function f(x) is continuous in 
the interval Xi'^x^x^ and if its derivative is always posi- 
tive (or always negative) in this interval, then the function is 
always increasing (or always decreasing) and so must have 
just one root between asj and x^. 

The root can be found approximately as follows. Consider 
the graph of the function 

2) y=m. 

Let 2/i =/(»!), Vi =f{x2), 

and draw the chord through the points (a^, y{) and (ajj, y^. 

The point in which this chord cuts the axis of x will obviously 
yield a further approximation to the root 
sought. Denote this last value by X. 
The equation of the chord is 

3) Jg - a^i ^ y - 2/1 ■ 

s^ — ^i 2/2 — 2/1 
Fig. 53 

On setting 2/ = and solving for x, we 

have, as the value of X, the following : 

4) X=xi-?i^^2,j, 

2/2-2/1 
or 

5) X=x,--^^^^^^f(x,). 




APPLICATIONS 171 

We have explained the method in detail and developed, in 
equations 4) and 5), the analytic formula for the determina- 
tion of the new approximation, X. In practice, however, it is 
usually simpler to draw the straight lines of Fig. 63 accurately 
on a generous scale and read off from the figure the value of X. 

Example. Consider equation 3) of § 2, Ex. 2 : 

6) a' + 2a; - 2 = 0. 
The curve in question is here 

7) y = ay' + 2x — 2, 

and the graphical solution of § 2 shows that the root is about 
a; = .7 or .8. 

Let a; = aji = .7 ; then j/i is found to have the value 

2/1 = -.267. 
Next, let x = X2 = .S; then y^ = .112. 
We have, then, to lay a secant through the points 
i^i, Vi) = (-7, - .257) and {x„ y,) = (.8, .112). 
Its equation is given by 3) * : 

a; -.7 ^ y + .257 
.8 -.7 .112 + .257' 

On setting y = in this equation and solving for x, we get, 

''.' X=.7+^^ = .7693. 

.369 

In order to see about how close this approximation is, com- 
pute the corresponding value of y : 

2/|z=.7693 = — -0063. 

We get, then, about two places of decimals, x = .77. 

* It is desirable that the student should make this determination 
graphically, as indicated above in the text. He should take 10 cm. to 
represent the interval of length .1, from Xi = .7 to aj = .8. 



172 CALCULUS 

It is possible to apply the method again, taking now 

(ao, 3/i)=(.7693, - .0063) 

and (a^, ^2) as before. We leave this as an exercise to the 
student. He should make both the graphical determination 
with an enlarged scale and the analytic determination of 
formula 4). 

The Method; Not, the Formula. The student may be 
tempted to use the formula 4) or 5), rather than to go back 
to the method by which it was derived. This would be un- 
fortunate, for the formula is not easily remembered, whereas 
the method, once appreciated, can never be forgotten. If the 
student finds himself in a lumber camp with nothing but the 
ordinary tables at hand, he may solve his equation if he has 
once laid hold of the method. It is true that the best way is 
for him to treat first the literal case and deduce the formula. 
But this he may not be able to do if he has relied on the 
formula in the book. 

EXERCISES 

Apply the method to a good number of the problems at the 
end of § 2. 

4. Newton's Method. Suppose again that it is a question of 
solving the equation 

1) f{x) = Q, 

and suppose we have already succeeded in finding a fairly 
good approximation, x = Xi. 

Consider the graph of the function 

2) y =/{<>=)■ 

Compute Hi =f(x,). To improve the approximation, draw the 
tangent at the point (xi, y^. Its equation is : 



APPLICATIONS 



173 



Evidently, this luie will cut the axis of a; at a poiat very near 
the point in which the curve 2) cuts this axis. If, then, we 
set y = in 3) and solve for x, we shall ob- y 
tain a second approximation to the root of 
1) which we seek. The value of this root 
will be 




4) 



X = xi- 



Fig. 54 






Example 1. Let us apply the method to the Example 
studied in § 3. In order, however, to have simpler numbers 
to work with, take Xi = .77 and compute the corresponding 



yi ; it is found to be : j/j : 



.0035. 



(«i,2/i) = (-77, -.0036). 
We must next compute dy/dx from the equation 
y = a^ + 2x—2; 



dx 



(dy\ =3.779. 



On substituting these values in 3), we have : 
2/ + .0035 =3.779 (a; -.77). 
Now set 2/ = and solve. The result is that given by 4) : 

.0035 



a; = .77 + 



3.779 " 



.7709. 



We have tabidated four figures in the result because this is 
about the degree of accuracy that seems likely. To test this 
point, compute y for the value of x which has been found : 

3/U=.7709 = --000L 

Since the slope of the graph is greater than unity, the error 
in a; is less than one unit in the fourth place. It is easy to 
verify the result by computing y for the next larger four-place 
value of a.: y |^..„,„ = + .0003. 



174 CALCULUS 

Thus we have a complete proof that the root lies between 
. 7709- and .7710, and we see that it lies about one quarter of 
the way from the first to the second value. 

Example 2. It is shown that the equation of the curve in 
which a chain hangs, — the Catenary, — is 

5) 2,=|^e^ + e-^ 

where a is a constant. The length of the arc, measujed from 
the vertex, is 

6) 






Let it be required to compute the dip in a chain 32 feet long, 
its ends being supported at the same level, 30 feet apart. 

We can determine the dip from 5) if we know a, and we 
can get the value of a from 6) by setting s = 16, x = 15: 



16 



Leta; = — . Then 



/ 15 15, 



f(x) = ei'-e-'-^x = 0, 

and we wish to know where the curve 

7) y=f{x) = e-e-'-^x 

crosses the axis of x. 
This curve starts from the origin and, since 

ax 

is negative for small values of x, the curve enters the fourth 
quadrant. Moreover, 

^=6^-6-- >0, x>0, 



APPLICATIONS 175 

and hence the graph is always concave upward. Einally, 

f(l) = e-e-^-2^ = .217>0, 

and so the equation has one and only one positive, root, and 
this root lies between and 1. 

It will probably be better to locate the root with somewhat 
greater accuracy before beginning to apply the above method. 
Let us compute, therefore, /(^). By the aid of Peirce's Tables 
we find : 

/(.5) = 1.6487 - .6066 - 1.0667 = - .0245 < 0. 

Comparing these two values of the function : 

/(.5) = -.02, /(I) = .22, 

and remembering that the curve is concave upward, so that 
the root is somewhat larger than the value obtained by direct 
interpolation (this value corresponding to the intersection of 
the chord with the axis of a;) we are led to choose as our first 
approximation Kj = .6 : 

/(.6) = 1.8221 - .5488 - 1.2800 = -.0067, 

/'(.6) = 1.8221 + .5488 - 2.1333 = .2376. 

Hence the value of the next approximation is 

X = .6 - z:j99K = .6 + .0282 = .628. 
.2376 

To get the next approximation we compute 

/(.628) = 1.8739 - .5337 - 1.3397 = .0005. 

Hence the value of the root to three significant figures is .628 
with a possible error of a unit or two in the last place, and the 
value of a we set out to compute is, therefore, 15/.628 = 23.9. 

Remark. Newton's method, like the other methods of this 
chapter, has the advantage that an error in computing the new 
approximation wUl not be propagated in later computations. 
Such an error wUl in general hinder us, because we are not 



176 CALCULUS 

likely to get so good an approximation. But the one test for 
the accuracy of the approximation is the accurate computa- 
tion of the corresponding y, and if this is done right, we see 
precisely how close we are to the desired root. 

The function /(») is usually simple, and it is easy to see 
whether the curve is concave upward or concave downward 
near the point where it crosses the axis. We thus have a 
means of improving the approximation at the same time that 
we simplify the new value of x. For, if the curve lies to the 
right of its chord, the approximation by interpolation will be 
too small ; and if the curve lies to the right of its tangent be- 
tween the point of tangency and the axis of x, the approxima- 
tion given by Newton's method will also be too small. 

Comparison of the Two Methods. When looked at from their 
geometric side the two methods appear much alike, the first 
seeming somewhat simpler, since it does not involve the use 
of derivatives. Why bother, then, with Newton's method? 
It is not a theoretical question, but purely one of convenience 
in carrying out the numerical work. It will be found that, as 
a rule, the first method is preferable in the early stages 
(usually, merely in the first stage). When, however, a fairly 
good approximation has been reached, the numerical work ra- 
volved in Newton's method is generally shorter than that 
required by interpolation. 

EXERCISES 

Apply the method to the Exercises of § 2. When, however, 
the approximation given by the graphical method of § 1 is 
crude, the method of interpolation may be used to improve it. 

5. Direct Use of the Tables. 

Example 1. Let us recur to the first example studied, Ex. 1, 
§2: 

1) cos a; = a?. 



APPLICATIONS 



177 



The graphical solution gave x = .75. Turn now to a table 
of natural cosines in radian measure, preferably Peiroe's Tables. 
As we run down the table, we find the entries : 



RADIANS 



.7389 
.7418 



cos NAT 



.7392 
.7373 



Thus X is seen to lie between .7389 and .7418. It is an ex- 
cellent exercise for the student to work out the interpolation 
for himself before we take it up at the end of the paragraph. 
The answer is : a; = .7391. 

Example 2. Consider the equation 
2) tan X = e', 

the desired root lying between and Tr/2. 

A free-hand drawing of the graphs of the functions 

y = tan x, y=e'' 

shows that x lies between 1 and 1.5. So the next step is taken 
conveniently by opening Peirce's Tables to the Trigonometric 
Functions and Huntington's to the Exponentials, and writing 
down the two pairs of values of the functions which came 
nearest together : 



1.3 
1.4 



tan X 

3.60 
5.80 



3.67 
4.06 



Thus the root is seen to lie between 1.3 and 1.4. 

The general case which the above examples are intended to 
illustrate is the following : — To solve thf equation 

f{x) = 4>{x), 

where f{x) and <^{x) are tabulated functions, or functions 
readily computed. 



178 



CALCULUS 



When the solution has progressed to the point indicated 
by the examples, the next step can be taken by interpolation, 
or by Newton's method, as will now be explained. 

Interpolation. When two values of the independent variar 
ble near together, Xi and x^, have been found such that /(») is 
greater than <^(a!) for one of them and less than ^(x) for the 
other, the best approximation to take next is the one given by 
the abscissa of the point of intersection of the chords of the 
graphs of the functions. 

This value, X, can be found as follows. 
Suppose that 

/(oh) < <^(a;i) and /(ajj) > (^(k^). 

Introduce the following notation : 

X — Xi^ h. 

From the figure, the triangles 
AiCBi and A2CB2 are similar, 
and 

AiBi = Ai, A^Bi = Aj. 

Their altitudes, when C is taken 
as the vertex, are respectively h 
and & — h. Hence 

h ^ B-h 
Ai Aj 

On solving this equation for h we find : 




3) 



A = - 



A, 



A1 + A2 



8. 



If /(a!i) > ^(xi) and/(a^) < ^(3:2), the result still holds, for 
Ai and A2 now become negative, but their numerical values 



APPLICATIONS 179 

correspond to the lengths of the sides of the triangles in 
question. 

It is easy to express in words the result embodied in 3). 

Etjle. In order to see what fraction of 8= x^ — Xi must be 
added to x^ in order to give X, form the differences 

Then the fraction is the quotient of the first of these differences by 
their sum. 

In practice, an accurately drawn figure on a large scale will 
often afford a quicker and sufficiently accurate solution. 

Example. Eetiiming to Ex. 1 above, we have : 
f(x) = cos X, ^(^)= ^ ; 

8 = X2-Xi = .0029, «! = .7389, % = .7418. 
4>(x{)-f{x0=-.OOO4: ; f{x^)- <t>(,x^)=- .0045. 

.0004_oo29 = :0116^_0002. 



.0049 49 

Hence the value of the new approximation is 
X = .7389 + .0002 = .7391. 

The student will have no difficulty in completing Ex. 2 above 
in a similar manner. It turns out that the correction is here 
less than one tenth of 8, and hence it does not influence the 
second place of decimals : x' = 1.30. 

Newton's Method. If a higher degree of accuracy is desired, 
it is well now to apply Newton's method to the function 

F{x)=f{x)- <l>{x). 

In the case of Ex. 1 above it is pretty clear that we already 
have four-plaee accuracy, and the computation of F{x) for the 
value X = .7391 would only verify the result. This is as far 
as we can go with four-place tables. If we needed greater 



180 CALCULUS 

accuracy, we should use Newton's method and five or six-place 
tables. 

Example 2 has been carried only to two-place accuracy, or 
three significant figures. We can obtain two further figures 
with the tables at our disposal, 

y = F{x)= tan x — e'. 

2/1 = F(1.30)= 3.602 - 3.669 = - .067. 



^ = sec^a;-e^ ^ 
dx dx 



= 13.97 - 3.67 = 10.30 



X=1.30-f — = 1.3067. 
10.3 

To test this result, however, would require five-place tables. 



EXERCISES 

Solve the following equations : 

1. cot a; = as, < a; < tt. 2. e* -|- loga!= 1. 

3. The hyperbolic sine (sh x or sinh a;) and cosine (ch x or 
cosh x) are defined as follows : 

sha5 = — - — , cha! = --L — , 

and are tabulated in Peirce's Tables, pp. 120-123. By means 
of these, reduce the treatment of Ex. 2, § 4, to the methods of 
the present paragraph. 

6. Successive Approximations. We come now to one of the 
most important of all the methods of numerical computation. 
In physics it is known as the method of Trial and Error ; in 
mathematics it goes under the name of the method of Succes- 
sive Approximations. 

The problem is that of solving a pair of simultaneous 
equations, 
1) F{x,y)=0, ^(x,y)=0. 



APPLICATIONS 



181 



The cases which arise in practice are characterized in general 
by t-wo things : First, there is only one solution of the equa- 
tions -which interests us, and the physical problem enables us 
to make a fairly good guess at it for the first approximation. 
Secondly, each of the equations 1) is simple, the curve can 
readily be plotted in character, and the equation can be solved 
with ease numerically for the dependent variable when a nu- 
merical value has been given to the independent variable. But 
elimination of one of the unknowns, though sometimes possible, 
is not expedient, since the resulting equation is hard to solve. 
The method is as follows. Plot the curves 1) in character 
with sufficient accuracy to determine which of them is steeper 
(i.e. has the numerically larger slope) at their point of inter- 
section. Let 



Ci: F(x,y)=0 

be the one that is less steep. 



or 



or 



y =/(«) 

X = (l>{y), 




y 


C>\ 










y^ 






\ 


■^l 




\ 


Hi 


y 


^ 












\ 


X 







a 


-2 ' 


"l 





Fig. 57 

the other. Then, making the best guess we can to start with, 
X = Xi, compute 2/1 from the equation of Cx : 

and substitute this value in the equation of G^, thus getting 
the second approximation: 

Proceeding with a^ in the same manner, we obtain first y^, 
then X3, and so on. 



182 CALCULUS 

The successive steps of the process are shown geometrically 
by the broken lines of the figures. 

The success of the method depends on the ease with which 
y can be determined when x is given in the case of Ci, while 
for Q^ X must be easily attainable from y. If the curves hap- 
pened to have slopes numerically equal but opposite in sign, 
the process would converge slowly or not at all. But in this 
case the arithmetic mean of x^ and x^ will obviously give a 
good approximation. 

The method has the advantage that each computation is 
independent of its predecessor. An error, therefore, while it 
may delay the computation, will not vitiate the result. 

Example. A beam 1 ft. thick is to be inserted in a panel 
10 X 15 ft. as shown in the figure. How long must the beam 
be made? 

We have : 

' sin 4> + l cos (^ = 15, 




15 [ cos ^ + 1 sin <^ = 10. 

Hence cos" ^ — sin" <^ = 10 cos <^ — 15 sin <^. 

Fig. 68 Now an expression of the form 
a cos ^ — 6 sin <^ 
can always be written as 

Va^ + V^( "■ cos ^ - ^ sin <^ ) = Va" + 6" cos (<t> + a), 
where cos a = - 



In the present case, then, 

cos 2<^=V325cos (<^ + a), 

, 10 . 15 
where cos a = — , sin « = — • 

V325 V325 

Thus a is an angle of the first quadrant and 
tan « = I, a = 56° 16'. 



APPLICATIONS 183 

Our problem may be formulated, then, as follows : To find 
the abscissa of the point of intersection of the curves : 

y = cos 2 <^, «/ = V326 cos (<^ + a). 

We know from the figure a good approximation to start 
with, namely : 

tan ^ = I, ^= 33° 44'. 

For this value of <^ the slopes are given by the equations : * 

i^ . ^ = _ 2 sin 2<^ = - 2 sin 67° 28' = - 1.8, 
IT d^ 

i^ . ^ = - V325 sin (<^ + «) = - V325 = - 18. 

Hence we have : 

Ci : y — cos 2 <f> ; 

On-, w = V326 cos (d> + a) or d) = cos^i — ^::rz — a- 

V325 

BegiiiniQg with the approximation 
^1 = 33° 44', 
we compute 2/1 = cos 67° 28' = .3832. 

Passing now to the curve Cj, we compute its <^ when its 

.3832 = V326 cos (<^2 + «), <^2 = 32° 31'. 

We now repeat the process, beginning with <j}2 = 32° 31' and 
fiiid : 2/2 = cos 65° 02' = .4221, 

.4221 = V325 cos (.^3 + a), <^3 = 32° 23'. 

A further repetition gives <^4 = 32° 22', and this is the value 
of the root we set out to determine. 

* Since the degree is here taken as the unit of angle, the formulas of 
differentiation involve the factor ir/lSO ; cf . Chap. V, § 2. 




184 CALCULUS 

EXERCISES 

1. Solve the same problem for a beam 2 ft. thick. 

2. A cord 1 ft. long has one end fastened at a point 2 ft. 
above a rough table, and the other end is 
tied to a rod 2 ft. long. How far can the 
rod be displaced from the vertical through 

— and still remain in equilibrium when 
released ? 
The equations on which the solution depends are : 

J 2 cot d + -= cot <j>, 

I 2 cos 5 + cos <^ = 2. 
If the coefficient of friction /a = ^, find the value of ^. 

3. A heavy ring can slide on a smooth vertical rod. To 
the ring is fastened a weightless cord of length 2 a, carrying an 
equal ring knotted at its middle point and having its further 
end made fast at a distance a from the rod. Find the position 
of equilibrium of the system. 

4. Solve Example 2, § 4, by the method of successive ap- 
proximations. 

7. Arrangement of the Xumerical Work in Tabular Form. 

In the foregoing paragraphs we have laid the chief stress on 
setting forth the great ideas which underlie these powerful 
methods of numerical computation. There are, however, cer- 
tain details of technique which are important, not only for 
ease in keeping in view the results obtained, but also for 
accuracy, since they reduce the numerical work to a system. 
We will illustrate what we mean by an example. 

Example. Let it be required to find all the values of x be- 
tween 0° and 360° which satisfy the equation 

sin a; = logio (1 — cos x). 



APPLICATIONS 



185 



A free-hand graph of each of the functions 

1) y = auix, y = logjo (1 — cos x) 

shows that there is one root between 
0° and 180° and a second between 
180° and 360°. But these roots 
cannot be located with any great 
accuracy iu this manner. It is nec- 
essary to do exact table work, and 
to keep the successive results in 
such form that they are convenient 
for later reference. 

To this end such a table as the following is useful.* 
with the trial value x = 150°. 




Fig. 60 



Begin 



X 


160° 










cosx 
1 — COS a; 
logio (1 — cos x) 


- .8660 

1.8660 

.2709 










sinx 


.5000 











Since the ordinate of the sine curve is larger than that of the 
logarithmic curve, it is clear from the figure that x is too small. 
Try X = 160°. 

Before proceeding further let us ask ourselves whether the 
above scheme is the simplest for the example in hand. Por 
the special value x = 160° we know cos x without reference to 
the tables, and hence one entry of the tables was sufficient. 
But when x = 160°, it will be necessary to enter the tables first 
for cos x, a second time for logio (1 — cos x), and still a third 
time for sin x. 



* Paper ruled in small squares is convenient for these tables, the in- 
dividual digits being written in separate squares. 



186 



CALCULUS 



1 — cos a; = 2 sin^-, 
2' 



logio (1 — cos x) = login ain^ - + logio 2 
= 2 log sin I +.3010. 

Hence it is possible to get along with, only two entries of tlie 
tables if we make use of tlie following scheme. 



X 


160° 


164° 


163° 3' 


\^ 


80° 


82° 


81° 32' 


logio sin ^ X 


1.9934 


T.9958 


1.9952 


2 logio sin \ X 


1.9868 


1.9916 


1.9904 


+ .3010 


.2878 


.2926 


.2914 


sin (180 - x) 


.3420 


.2756 


.2916 



The ordinate of the sine curve is still in excess, but only 
slightly so. Try x ~ 164°. It is seen that the curves have 
now crossed. Moreover, the two approximations for x — 
namely, 160° and 164° — are so near together that we can with 
ad-vantage apply the method of interpolation of § 5. We 

■'^ '■ /(o!) = logio (1 — cos a;) ; 



^ (a;) = sin x, 
x^ = 160°, 
c/,(a:!i) = .3420, 
<^(k.^) = .2766, 

7i = 



Ai 



Xi = 164°, 
f{x^) = .2878, 
/(a^) = .2926, 

» .0542 , 



i = 4°; 
Ai = .0542 ; 
A, = .0170; 



A1 + A2 



.0712 



: 3.06. 



Thus the correction is seen to be 3.05°, or 3° 3', and 
the new approximation is : 

a; =163° 3'. 



APPLICATIONS 187 

For this value of x the values of the two functions, /(as) and 
<^ (x), differ by a quantity -which is comparable with the error 
of the tables, and the problem is solved. 

EXERCISES 

1. Determine the other root in the above problem. 

3. Solve the equation : 

cot X = login (1 + sin a;), < a; < 90°. 

3. Find the positive root of the equation 

e~' = 0? — X. 

Suggestion. Tabulate x, a? (from a table of cubes), a;' — x, 
and e""'. 

8. Algebraic Equations. By an algebraic equation is meant 
an equation of the form 

1) aoa;" + aiX^-^ + - + a„ = 0, ao¥= 0, 

where n denotes a positive integer. 

If the coefBcients Oq, aj, •■■ are numerical, the roots can be 
approximated to by the method of interpolation or by" Newton's 
method. In either case it becomes necessary to compute the 
value of the polynomial 

f(x) = OqW + aia;»-i + •■ •+ a„ 

for several values of x, the later ones of which will be at least 
three- or four-place numbers. There are labor-saving devices 
for performing these computations, to which we now turn. 

Numerical Computation of Polynomials. Let a cubic poly- 
nomial, for example, be given : 

f(x) = ax^ + bx^ + cx-\-d, 

and let it be required to compute f{x) for the value x = m. 
Write down the following scheme : 

_a am -\- b am^ + bm-\-c f(m) 

am, am^ + bm am? -\- hm? + cm 



188 CALCULUS 

the explanation of wMeh is as follows. Begin with the first 
coeflB.cient, a, and multiply it by m to get the expression am 
which stands below the line. To this expression add the 
second coefficient, 6, to get the second expression above the 
line, am + b. Next, multiply this expression by m to get 
the expression which stands below it, and continue the process. 
The last entry above the line will be the required value, 

/(m) = aw? + bm? + <mi + d. 

Example. Let 

f{x) = 7a? -&X-- + Zx -1, 

and let it be required to compute the value of /(«) for x = .8. 
Here, the scheme is as follows : 

7 -.4 2.68 -4.856 

5.6 -.32 2.144 
and hence 

/(.8) = - 4.866. 

It will be observed that the process requires only additions 
(or subtractions) and multiplications. The former can be per- 
formed mentally. The latter are executed most simply by 
one of the machines now in general use with computers. 
These instruments, combined with the method of this para- 
graph, have rendered Horner's method for solving numerical 
algebraic equations obsolete. 

EXERCISE 

Compute the value of 

5.1a^ - 3.42a;'- -H 1.432a; -|- .8543 
for X = .1876. 

In the problems which arise in physics, however, it is not 
a question of computing all the roots of a numerical equation, 
about which nothing is known beyond the coefficients. Usu- 
ally, the equation is a cubic or biquadratic, and only one root 
is required. Moreover, from the nature of the problem, a close 



APPLICATIONS 189 

guess at the value of this root can be made at the outset. 

Then the methods set forth in this paragraph and in §§ 2, 3 
lead quickly to the desired result. 



EXERCISES 

Solve the following equations, being given that there is one 
root, and only one, between 0° and 90° : 

1. 4 cos3 (9 - 3 cos ^ = .6283, 0° < 6 < 90°. 

2. sin' e - .75 sin $ = .1278, 0° < (9 < 90° 

3. Find the root of the equation 

!B* + 2.6SC3 - 5.2a;2 - 10.4a; + 5.0 = 
which lies between and 1. 

4. Find the root of the equation 

3x* - 12a? + 12a;2 -4 = 
which lies between 2 and 3. 

9. Continuation. Cubics and Biquadratics. Aside from the 
special problem of numerical computation, the simpler alge- 
braic equations present an intrinsic interest which should not 
be ignored. 

Transformations, a) Let the cubic equation 

1) /(») = aa? + bx'^ + cx + d = 0, a^O, 
be given, and let x be replaced by y, where 

2) y = x — h, x = y -\-h. 

Then 

f(x) = a(y + hy + b(y + hf J^c{y + h) + d = ^{y) 
= af + (3ah + b)y''+-, 

where the later coefficients are easily written down. 



190 CALCULUS 

If y = )8 is a root of the equation 
3) <^(y) = o, 

then x = p + h 

•will he a root of equation 1). For, it is always true that 

/{"') = Hy) 

when X and y are connected hy the relation 2). 

Here, h is any number we please. In particidar, h can 
always he so chosen that the coefficient of the second term of 

3) will drop out. It is sufficient to set 

4) 3ah + b = 0, or A = -A. 

Obviously, the same method can be used to transform an 
algebraic equation of any degree into a new equation whose 
second term is lacking. 

EXERCISES 

Transform the following equations into equations in which 
the second term is lacking. 

1. a.-3 + a;'- — a; + 1 = 0. 2. So? — ix'^ +2 = 0. 

3. a;* + a^ — a;2 + l = 0. 4. 5a^ —ia^ + x^ +x — SO=zO. 

5. 3x^—7ai' + x^ — x — l = 0. 6. a;6 + a^ + iB2^a,.^l_0. 

6) Let the equation 

5) f(^x)=x>+pa? + qx + r=0 
be given, and let x be replaced by y, where 

6) y = ^, x=ky. 

Then 

f(x) = kY + k-^ + kqy + r. 

Denoting this last polynomial by <^ (y), we have 

fip) = 'f>{y) 

for all values of x and y which are connected by the relation 6). 



APPLICATIONS 191 

It is clear that, if ^ = j8 is a root of the equation 

then x = kp will be a root of 5). 

The factor k is arbitrary, and we can always determine it so 
that, on dividing equation 7) through by fc : 

the coefficient of j/' will be numerically equal to unity (provided 
that p^O): 

i) ^=1 or k = \/p, if p>0; 

ii) =2 = — 1 or k = ^/—p, if p <0. 

k^ 

In this way, equation 5) can be reduced to one of the two 
forms 

a) y* + y^ + Ay + B = 0; 

/8) y*-y^ + Ay + B = 0. 

If, in particular, p = and q ^0, 5) can be reduced to the 
form 

y) y' + y + B = 0. 

The method can be applied to any algebraic equation whose 
second term is lacking : 

x" + CiX"-^ + CsK"-' 4- - + c„ = 0. 



EXERCISES 

1, Replace the equation 

7a!^ - 175a;2 + 16a; + 10 = 

by an equation of the type j8), and state precisely the relation 
of the roots of the second equation to those of the first. 



192 Calculus 

2. Show that, if in the equation 

Ooa" + aiX°~^ + — + a„ = 0, 
where Oo =?^ and a„ =^ 0, the transformation 

1 

2/ = - 

X 

is made, the roots of the new equation, 

a^" + «„-i3r"' H 1- oo = 

are the reciprocals of the roots of the given equation, 

3. If on transforming equation 1) by 2), where h is deter- 
mined by 4), the constant term in the resulting equation 3), 
<^(y) = 0, does not vanish, the further transformation 

8) ' y = -, or x = - + h, 

z z 

will carry 1) into an equation ia which the linear term is lack- 
^S- Az^ + B^ + D=:0. A^O, D=^Q. 

The theorem holds in full generality for an algebraic equa- 
tion of any higher degree. State it accurately. 

4. Replace the equation 

a^ — 4:a^ — 6af + 16x-4: = 
by an equation of the type 

Ay* + Bf+Of' + D = 0. 

Graphical Treatment. We have already seen that the cubic 
a^+px + q = 
can be solved graphically by cuttiag the standard graph 

y = ai> 

by the straight liae, 
•' ° y = -px-q. 

Since the general cubic can be reduced by the transformation 
2) to a cubic of this type, we may consider the general problem 
of the graphical solution of a cubic as solved. 



APPLICATIONS 193 

To obtain a similar solution for tlie general biquadratic, 
9) aoil^ + ba^ + ex"- + dx + e = 0, a 4=0, 

begin by reducing it to one of the three forms : 
i) y* + y^ + Ay + B = 0; 

ii) y*-y2 + Ay + B = 0; 

iii) y' +Ay + B = 0. 

An equation of type i) : 

a^ + x'-+ Ax + B = 0, 
can be solved graphically by cutting the standard curve 

y = !e^ 
by the parabola 

y =— x' — Ax — B. 

A similar procedure leads to a solution in the case of each 
of the other two types, ii) and iii). 

The Method of Curve Plotting. Let the coefficients a, e in 
equation 9) be different from 0. By means of Ex. 3, p. 192, the 
equation can be reduced to one of the following type : 

J[a;« + B3? + Cx^ + E = (i. 

In order to discuss the number and location of the roots of 
this equation, it is sufficient to plot the curve 

y = Aa^ + Ba?+ Cx^+E. 

Since all the maxima, minima, and points of inflection of this 
curve can be determined by means, at most, of quadratic equa- 
tions, the problem is readily solved in any given numerical case. 

EXERCISES 

Determine the number of real roots of each of the following 
equations, and locate them approximately. 

1. 3a!* + 8a;3 -90a:2+ 100 = 0. 

2. Sa;* + 8a;' -90a;5 + 600 = 0, 



194 CALCULUS 

3. 3a!* + 8 a? -90aii2 + 1600 = 0. 

4. Show that the eqiiation 

has rlo real roots. 

How many real roots has each of the following equations ? 

5. a;^— 5x — 1 = 0. 6. a;' + 7a;— 1=0. 
7. a? — 4a; + l=0. 8. a;' -3a; — 2 = 0. 

9. a;'-x + 3 = 0. 10. 43;=- 15a;2 + 12a;+ 1 = 0. 

11. 3a;< + 4a;' + 6a!2-l = 0. 12. 3a;< - 4a?' + 12a;= + 7 = 0. 

13. How many positive roots has the equation 

6a;* + 8a;3 - 12a;2 - 24a; - 1 = ? 

14. Has the equation 

3a;»_8a;«+12ar' + l = 
any real roots ? 

16. By means of the graph of the function 

y = a?+px + q 
show that the equation 

a? +px + q = 
has 

(a) 1 real root when ^ + — > ; 
^7 4 

(6) 3 real roots when ^ + 2. < o ; 
^ ^ 27 4 ' 

(c) 2 real roots when ^ + i- = 0, {p and q not both } 

fd) 1 real root when ^ + 21 = 0, {p = g = 0} 
^ ' 27 4 

In case (c) it is customary to count one of the roots twice ; 
in case (d), to count the root three times. 



APPLICATIONS 



195 



16. Extend the criterion of Ex. 15 to the case of the general 
°^^^° aa:? + bofi + cx + d = 0. 

10. Curve Plotting. We will close this chapter by consider- 
ing the application Of the principles set forth in the earlier 
paragraph-on curve plotting (Chap. Ill, § 6) to some interest- 
ing curves of a more complex nature. 



1) 



Example 1. To plot the curve 

1 1 



y- 



l''"a!-|-l 



The curve is obviously not symmetric in either axis ; but 
the test for symmetry in the origin is fulfilled, since on replac- 
ing a; by — K and yhj—y the new equation, 

1 . 1 



-y = 



X — 1 — X + 1 



is equivalent to the original equation, 1). Incidentally we 
observe that the curve passes through the origin. 

In consequence of the symmetry just noted it will be 
suffi.cient to plot the curve for positive values of x and then 
rotate the figure about the origin through 180°. 

To each positive 
value of X but one 
there corresponds 
one value of y. 
When X approaches 
1 as its limit from 
above (i.e. always 
remaining greater 
than 1), y becomes 
positively infinite. 
Hence the line x=l 
is an asymptote for 
one branch of the 



curve. 



FlQ. 61 



196 CALCULUS 

When X approaches 1 from below, y becomes negatively 
infinite, and hence this same line, x=l, is an asymptote for a 
second branch of the curve. 

Tor all other positive values of x, y is continuous. 

The slope of the curve is given by the equation 



2) ^ = -(. 

' dx V' 



{x-iy (a; + 1)2/ 

and is seen to be negative for all values of x for which y is 
continuous. Thus, ia particular, the curve is seen to have no 
maxima or minima, or in fact any points at which the tangent 
is horizontal. 

The second derivative is given by the formula 

' dx'' \{x - Vf (x + 1)V 

When a; > 1, the right-hand side of this equation is always 
positive, and so the curve is concave upward in this interval. 
Moreover, it is evident from 1) that, when a; = + oo, y ap- 
proaches from above, and so the positive axis of x is also an 
asymptote. 

In the interval < a; < 1, the second derivative is surely 
sometimes negative, for this is obviously the case when x 
is only slightly less than 1. Is dhj/dx^ always negative in 
this iaterval? If not, it must pass through the value 0; 
for a continuous function cannot change from a positive to 
a negative value without taking on the intermediate value 
0.* Let us set, then, the right-hand side of equation 3) 
equal to and solve: 



2f— L_ + _^_Vo. 



* How must the graph of a, continuous function look, which is' some- 
times positive and sometimes negative ? It must cross the axis of ah- 
scissas, must it not ? At the point or points where It crosses, the function 
has the value 0. 



APPLICATIONS 197 

This equation is equivalent to the following : 
1 ^ 1 

(x-iy (x + iy' 

Extracting the cube root of each side of this equation, we have : 

1 1_ ■ 

x—1 x+1 

Clearing of fractions we find : 

x + l=-(x-l), 
or 2a! = 0. 

Hence a; = is the only value of a! for which (Py/dx'^ can 
vanish, and we see at once that the right-hand side of 3) does 
vanish for a! = 0. 

We have thus proven that the contiauous function 3) is no- 
where iu the interval < a! <.l, and since it is negative in 
part of this interval, it is negative throughout. Hence the 
curve is concave downward throughout the interval. 

It is now easy to complete the graph. The curve has one 
point of inflection, — namely, the origin, — and the slope there 
is, by 2), equal to — 2. 

EXERCISES 



Plot the following curves : 






'■ y-,U- 


2. 


3a! 

^ 3-|-a:2 


'■ ^-a!-2 + a! + 2- 


4. 


y=-+ \ 

X x—1 


'■ y-x + x + 1 


6. 


1 


y x^-1 


7. y=^- 

x^ 


8. 


V- ^ 


(i-xy 


1 


10. 


V- ^ 


^- ^-a!'" 


(x + iy 



198 CALCULUS 

11. y = x-\ 12. y = X 

X X 

A. ^ 

13. y = —^ + 3? — 2x. 14. y = - .&x — a?. 

1 — a; d +» 

15.,=^-^. 16. , = 1— A_. 

a;— la; + l x x—1 

Example 2. To plot the curve 
4) y^ = x^ + a?. 

We observe first of all that the curve is symmetric in the 
axis of a;. It is sufficient, therefore, to plot the e^rve for posi- 
tive values of y, and then fold this part of the curve over on 
the axis of x. The curve goes through the origin. 

Unlike the examples hitherto considered, this curve does 
not permit ah arbitrary choice of x. It is only when the right- 
hand side is positive or zero, i.e. when 

a!2 + a? ^ 0, 
or 

a?(l + x)>0 or x>—l, 

that there will be a corresponding value of y and thus a point 
with the given abscissa. 

Obviously, the curve cuts the axis of x at the origin and at 
the point x = — l. We have, then, essentially two problems : 

i) to plot the curve for a; > ; 
ii) to plot the curve for — 1 < a; < 0. 

i) When a; > 0, the positive value of y Is given by the 
equation 



6) y = xVT+x. 

Hence 

dy^ 2 + 3a; _ 
dx 2Vl-|-a; 



APPLICATIONS 



199 



For positive values of x the right-hand side of this equation 
is always positive, and hence there are no horizontal tangents 
in the interval under consideration ; the slope of this part of 
the curve is always positive. In particular, the slope at the 
origin is unity : 



dx 



= 1. 



7) 



The second derivative has the value 

d^y _ 4 + 3a; 
<*'«' 4(1 + x)^ 



The right-hand side of this equation is always 
positive in this interval, and thus it appears 
that the curve is concave upward for all posi- ° 
tive values of x. 




Fig. 62 



ii) When — 1 < a; < 0, the positive value of y is no longer 
given by the formula 6), since x is now negative.* In the 
present case, 



8) 


y=~ a; VI + x, 


and consequently 
9) 


dy _ 2-1- 3a; 
dx 2V1 -t- X 


io-> 


oPy_ 4-1- 3a; 



i^^"' 4(1 + x)i 

The first derivative will vanish if, and only if, 

2-|-3a; = 0, 

or x = — ^. 

* The student must have clearly hi mind the definition of the function 
expressed hj the y sign, which was laid down in Chap. I, § 1. This func- 
tion is the positive square root of the radicand ; It can never take on a 
negative value. 



200 



CALCULUS 



It is, therefore, important to determine the corresponding 
point on the curve and draw the tangent there : 

2/|x=-!=-(-|)V5rri=^=.38. 

Two other important poiots for the present curve are the 
origin and the point a; = — 1, y = 0. At these 
points the slope has the following values : 




-1 .1. 



dx 



— 1; 



dy 
dx 



Fto. 63 Draw the corresponding tangents. 

From the expression 10) for the second derivative it is 
clear that, when — 1 < a; < 0, the right-hand side of this 
equation is always negative, and 
hence the curve is concave down- 
ward throughout the whole in- 
terval in question. We can now 
draw in the curve in this interval, 
Kg. 63. 

The curve is now complete above 
the axis of x. It remains, therefore, 
merely to fold this part over on that 
axis. The entire curve is shown in 
Fig. 64. 

EXERCISES 




Fio. 64 



Plot the following curves : 

1. y^ = x^-a?. 2. y^ = x- Ix"- -|- 7?. 

3. y'i={x — a)\Ax-\-B) 

Suggestion : Write the second factor in the form 

Jx + B = A(x — b), where b = —, 

and make two cases : i) A>0; ii) A<0. Discuss the omitted 
case, ^ = 0. 



APPLICATIONS 201 

4. 2/2 = a;2 - »*. 5. yi — y^^ jgi. 

Example 3. To plot the curve 

11) y'' = x{x-l){x-2). 

The curve lies wholly in the regions 

^ a) ^ 1 and 2 ^ sc. 

It is symmetric in the axis of x, and hence it is sufficient to 
plot it for positive values of y. 
The function 

y =:-yJx(x — l)(x — 2) 

is continuous in the interval ^ a; ^ 1. It starts with the 
value when x = 0, increases, and finally decreases to when 
a;=l. 

When X, starting with the " 
value 2, increases, y, starting 
with the value 0, increases, 
always remaining positive, and 
increasing without limit as x be- 
comes infinite. 

So much from considerations of continuity. A piore specific 
discussion of the character of the curve can be given by means 
of the derivatives of the function. 

The slope is given by the formula 



Fia. 65 



12) 

or 

13) 



2y^=Zx^-&x + 2 
dx 

dy ^ Zx^—Qx + 2 
dx ~ 2^x{x-l){x-2) 



The slope is infinite when a; = or 1 : 

= co. 



dy 
dx 



dy 
dx 



At these points, the tangent is vertical. 



202 CALCULUS 

The slope is when 

3a;2-6a!+2 = 0. 
The roots of this equation are 

x=l+—, x = l-—- 

V3 V3 

The first of these values does not correspond to any point on 
the curve. The second, x = .42, yields a horizontal tangent, 
the ordinate being 



" \3V3 



3V3 

Plot this point and draw the tangent. From the above dis- 
cussion on the basis of continuity it is obvious that this point 
must'lbe a maximum, and we see that there 
are no other maxima or minima. But it 
is not clear that the curve has no points of 



inflection in this interval. 
iG. bb . rpg treat this question, compute the sec- 

ond derivative. This might be done by means of formula 13) ; 
but it is simpler to use 12) : 

2y^ + 2^=6x-6, 

y^ = 3x-3-^. 
"dx^ dx^ 

Substitute here the value of dy/dx from 13) and reduce : 

■' '^dx"' 4 a; (a; -1) (a; -2) 

And now we seem to be in difficulty. How are we going to 
tell when dh//dx^ is positive, when negative ? 

First of all, y is positive, and so the sign of dhjjd^ will be 
the same as that of the right-hand side of the equation. 

Secondly, in the interval in question, < a; < 1, the denomi- 
nator is positive. 



APPLICATIONS 203 

All turns, then, on whether the numerator, i.e. the funotion 

15) w = 3!B*-12a!8 + 12a;2-4, 

is positive or negative. To answer this question, plot the 
graph of the function 15). The slope of the graph is given by 
the equation 

16) ~ = 12a^ - 36a;! + 24a! = 12 a; (a; - 1) (a; - 2). 
dx 

In the interval in question, the right-hand side of this last 
equation is always positive. Hence u increases with x through- 
out the interval ^ a; ^ 1, and consequently attains its great- 
est value at the end-point, x = 1. Here, 

Mi»=l = -1- 

We see, therefore, that u is negative 
throughout the whole interval in question, 
and consequently the graph of 1) is Concave 
downward in this interval. 

The reasoning by which we determined whether u is pos- 
itive or negative is an excellent illustration of the practical 
application of the methods of curve plotting which we have 
learned. It is in no wise a question of the precise values of 
u which correspond to x. The question is merely : Is u posi- 
tive, or is it negative? Without the labor of a single com- 
putation involving table work we have answered this ques- 
tion with the greatest ease. Such questions as these arise 
again and again in physics, and the aid which the calculus is 
able to render here is most important. 

One further point. It may seem to have been a fluke that 
we were able to factor the polynomial in 16) and thus simplify 
so materially the further discussion. And yet, in the problems 
which arise in practice,' — the problems with a pedigree, — just 
such simplifications as this present themselves with great 
frequency. 




FlQ. 67 



204 



CALCULUS 



To complete the graph, it remains to consider the interval 
2 ^ a; < oo. Since . 

the tangent to the curve is vertical at the point where the curve 
meets the axis of x. It is clear, then, that the curve must be 
concave downward for a while, and so dPy/da? < for values 
of X slightly greater than 2. This is verified from 14), since 

17) m|^=-4. 

On the other hand, when x is large, u is positive and (Py/da^ 
is positive. Hence the curve is concave upward. There must 
be, therefore, a point of inflection in the interval, and there 
may be several. 

From 14) we see that the second derivative will vanish when 
and only when 3^ _ jgj^ + i2a,2 -4 = 0. 

The problem is, then, to determine the number of roots of 
this equation which are greater than 2, and to compute them. 

Again, it is a question of 
the graph of 15). When 
« > 2, we see from 16) 
that 

> 0. 




du 
dx 



Hence u steadily increases 
with X. Now, from 17), 
u starts with a negative 
value, and u is positive 
and large when x is large. 
Hence u vanishes for just 
one value of x which is 
greater than 2. Since 
m|j_3 = 23, this root is seen 
to lie between 2 and 3. 
It can be determined to 
any required degree of accuracy by the foregoing methods of 



Fig. 68 



APPLICATIONS 205 

this chapter, which find herewith a practical application. To 
two places of decimals it is 2.47. 

EXERCISES 

Plot the following curves ; 

1. y = a^ — X. 2. y = x — a?. 

3. 2/2 = ajs Ij. X. 4.' 2/"! = 1 — ar*. 

5.^2/2=(a!2 — l)(a;2-4). 6. y"^ = {1 - x'^){x''- — i). 

2 = - « «2 = i 





^ OI?-X 


9. 


7/' '^ . 


y -1-x 


w 


V'- ="' 




•' 1 + x^' 


13. 


V'- '""' 


^ x-1 


15. 


y-^ = a? — A:X'' + 5x. 


17. 


2/ = sin a;— sin 2a!. 


19. 


y == cos a — cos 2a!. 



8. 2/2 = 
10. y 



(a;2 - l)(a!2 - 4) 

X 



1+x 
12- t=^^ 



14. 2/' = T^- 
1 + a; 

16. 2/ = sin a; + sin 2 a:. 

18. y = cos X + cos 2a!. 

20. y = x-\- sin a!, ^ a! ^ tt. 



CHAPTER VIII 
THE INVERSE TRIGONOMETRIC FUNCTIONS 
1. Inverse Functions. Let 

(1) y^m 

be a given function of x, and let us solve this equation for x as 
a function of y : 

(2) '^ = 't>(y)- 

Then <j}(y) is called the inverse function, or the inverse of the 
function /(k). Thus if /(«) = a^, we have 

y = af'. 

Hence x ^ Vy, 

and <^(2/) is here the function -y^ 

When the given function is tabulated, the table also serves 
as a tabulation of the inverse function. It is necessary merely 
to enter it from the opposite direction. Thus, if we have a 
table of cubes, we can use it to find cube roots by simply re- 
versing the roles of the two columns. 

In the same way, the graph of the fxmction (1) serves as the 
graph of the function (2), provided in the latter case we take 
y as the independent variable, and x as the dependent variable, 
or function. 

The graph of the inverse function, plotted with x as the in- 
dependent variable, can be obtained from the former graph as 
follows. Make the transformation of the plane which is de- 
fined by the equations : 

(3) . ^ = ^'1 or ''=y''\ 

206 



THE INVERSE TRIGONOMETRIC FUNCTIONS 207 




Fig. 



It is easy to interpret this transformation. Any poipt, whose 
coordinates are (x, y), is carried over into a point {x', y') situ- 
ated as follows : Draw a line L tlirough the origin bisecting the 
angle between the positive 
axes of coordinates. Drop 
a perpendicular from {x, y) 
on L and produce it to an 
equal distance on the other 
side of L. The poiat thus 
determined is the point 
(a;', y'). The proof of this 
statement is immediately 
evident from the figure. 

Thus it appears that the 
transformation (3) can be 
generated by rotating the 
plane about L through 
180°. 

The transformation is also spoken of as a reflection in L, 
since if a plane mirror were set at right angles to the plane of 
(x, y) and so that the line L would lie iu the surface of the 
mirror, the image of any figure, as seen in the mirror, would 
be the transformed figure. 

Monotonic Functions. A function, f(x), is said to be mono- 
tonic if it is single-valued and if, as x increases, /(x) always 
increases, or else always decreases. We shall be concerned only 
with functions which are, in general, continuous. It is obvious 
that the inverse of a monotonic function is also monotonic. 

A given function, 

2/ =/(»'), 
can in general be considered as made up of a number of pieces, 
each of which is monotonic in a certain interval.* Thus the 
function 
(4) y = sfi 

* There are functions which do not have this property ; but they do 
not play an important rdle in the elements of the Calculus. 



208 CALCULUS 

can be taken as made up of two pieces, corresponding respec- 
tively to those portions of the graph ■which lie in the first and 
the second quadrants, the corresponding intervals for x being 
liere -oo<a;<0, 0<a;<oo. 

Each of the pieces, of which /(cc) is made up, has a monotonic 
inverse, and thus the function ^(a;) inverse to f(x) is repre- 
sented by a number of monotonic functions. 

In the example just cited, the inverse function is multiple- 
valued : 

(5) y=±-Vx. 

But one of the two pieces into which the original function was 
divided yields the single-valued function 

(6) y=V5, 

the so-called principal value of the multiple-valued function 
(5); the other, y = _V^, 

the remainder of (6). 

The derivative of a monotonic function cannot change sign ; 
but it can vanish or become infinite at special points. Thus 

y = Va^ — a;2, ^ a; ^ a, 

is a decreasing monotonic function. Its derivative is, in gen- 
eral, negative ; but when a; = 0, it vanishes, and- when x=a, 
it becomes infinite. 

Differentiation of an Inverse Function. The function <f>{x) 
inverse to a given function f(x) can be differentiated as follows. 
By definition, the two equations 

(7) y=^{x) and x=f(y) 

are equivalent ; they are two forms of one and the same relar 

tion between the variables x and y. Their graphs are identical. 

Take the difEerential of each side of the second equation : 

dx = d/{y) = DJiy).dy. 



THE INVERSE TRIGONOMETRIC FUNCTIONS 209 



Hence 
(8) 



dx 



DJ{y) 

To complete the formula, express the right-hand side of (8) 
in terms of x by means of (7). 

2. The Inverse Trigonometric Functions. The inverse trigo- 
nometric functions are ■ chiefly important because of their 
application in the Integral Calculus. They are defined as 
follows. 

(a) The Function sin~i x. The inverse of the function 

(1) • y = sin X 

is obtained as explained in § 1 by solving this equation for x 
as a function of y, and is written : 

(!') X = sin~i y, 

read " the anti-sine of y." * In order to 
obtain the graph of the function 

(2) y-=sm~^x 

■we have, then, merely to reflect the graph 
of (1) in the bisector of the angle made 
by the positive coordinate axes. We 
are thus led to a multiple-valued func- 
tion, since the line x = x'(— 1 ^ a;' ^ 1) 
cuts the graph in more than one point, 
— in fact, in an infinite number of points. 
For most purposes of the Calculus, how- 
ever, it is allowable and advisable to pick 

* The usual notation^on the Continent for sin-i a;, tan-i x, etc., is arc sin x, 
arc tan x, etc. It is clumsy, and Is followed for a purely academic reason ; 
namely, that sin-' x might he misunderstood as meaning the minus first 
power of sin x. It is seldom that one has occasion to write the recipro- 
cal of sin X in terms of a negative exponent. When qne wishes to do so, 
all ambiguity can he avoided by writing (sin x)-K 




210 CALCULUS 

out just one value of the function (2), most simply the value 
that lies between y = — t/2 and y = + ■rr/2, and to understand 
by sin~'a! the single-valued function thus obtained. This de- 
termination is called the 2>rincipal value of the multiple-valued 
function sin"' a;. Its graph is the portion of the curve in 
Fig. 70 that is marked by a heavy line. This shall be our 
convention, then, in the future unless the contrary is explic- 
itly stated, and thus 

(3) y = sin"' x 

is equivalent to the relations : 

(30 a;=siny, -2^2/<|- 

In particular, 

sin-iO = 0, sin-il=^, sin-i(- 1) = -|. 

The student should now prepare a second plate, showing 
the graphs of the three functions sin"' x, cos~' as, tan"' w. Place 
the first in the upper left-hand corner of the sheet ; the second, 
in the upper right-hand corner ; and the third on the lower 
half-sheet. All of these curves can be ruled from the templets. 
Use a fine lead-pencil ; then mark in the principal value of 
the function in a clean, firm red line. Also mark, in each fig- 
ure, all the principal points, as is done in Fig. 70 of the text. 

Differentiation of sin"' x. In order to differentiate the func- 
tion . _, 

y = sm ' X, 

make the equivalent equation, 

a; = sin 3/ 

the point of departure. Then 

dx = d sin y = cos y dy. 

Hence ' ^ ^. 

dx cosy 



THE INVERSE TRIGONOMETRIC FUNCTIONS 211 



Tlie right-hand side of this eqiiation. cau be expressed in 
terms of x as follows. Since 

sin2 y + C032 y = l 

and since sin y = x, we have , 

cos' y = 1 — x% cos y = ± Vl — x''. 

We have agreed, however, to understand by sin~ia! the 
principal value of this function. Hence y is subject to the 
restriction : — ir/2 ^y^ ■jr/2, and consequently cos y is posi- 
tive (or zero). We must, therefore, take- the upper sign before 
the radical,* the final result thus being : 



(4) 



d . _, 1 

— sin 1 X = — 

dx Vl- 



(4') 



d sin~i X = 



dx 



Vl-a? 



(b) The Function cos~ia;. The treatment here is precisely 
similar. The definition is as follows : 

(5) y = cos~i X if x = cos y, 

(read : " anti-cosine x "). 

The graph of the function cos~ia; is as 
shown in Fig. 71. Like sin^ia;, this function 
is also infinitely multiple-valued. A single- 
valued branch is selected by imposing the 
further condition 

O^y^ir. 

This determination is known as the principal 
value of cos~i x : 



(6) 



y- 



O^y^TT. 




* Geometrically the slope of the portion of the graph in question is 
always positive, and so we must use the positive square root of 1 — x''. 



212 CALCULUS 

It will be understood lieiiceforth. that the principal value is 
meant unless the contrary is explicitly stated. 

In preparing the graph of this function, mark the principal 
value as a firm red line. 

To differentiate the function cos~ia!, use the implicit form 
of equation (5) : 

X = cos y. 
Hence 

dx = d cos y = — sin y dy 



3>iid. 




dj _ 
dx 


1 






siny 




For the 


principal 


value, sin^ 


is positive. 


and hence 


(V 




d 

— cos~ia; = 
dx 


1 , 




Vl-a;2 




or 










(7') 




d cos~i X = 


dx 





vr 

The principal values of the functions sin~* x and cos~i x are 
connected by the identical relation : 

(8) sin~i9!+ cos~ia; = ^- 

By means of this relation, the differentiation of cos"* a 
could have been performed immediately. 

(c) The Function tan"* x. Here, the definition is as follows : 

(9) 2/ = tan~ia; if a; = tany, 

(read : " anti-tangent x "). 

The principal value is defined as that determination which 
lies between — Tr/2 and Tr/2 : 

(10) 2/ = tan-ia!, _|<2,<|. 

In preparing the graph of this function, mark the principal 
value as a firm red line. 



THE INVERSE TRIGONOMETRIC FUNCTIONS 213 





y 


^ . 






2 







/^ 03 


^ 






T^ 






3 





Fia. 72 

To differentiate tan"** use the implicit form (9). Hence 

da; = d tan y = sec^ y dy, 

dy^ 1 

da: sec^ y 

Since sec^ y = 1-\- tan^ y 

and tan y = x, it follows that 

(11) 
or 

(12) d tan-la; = -^. 
^ ' 1 4- a;'' 

(d) 2%e Function cot-i a;. Here, the definition is : 

(13) y = cot-i a; if x = cot y, 

(read : " anti-cotangent x "). 

The principal value is chosen as that one which lies between 
and TT ; 

(14) y >= cot-' X, < y < TT. 



— tan-i X = , 

da; 1 + a;2' 



214 CALCULUS 

The differentiation can be performed as in the case of the 
function tan"ia;, but still more simply by means of the identi- 
cal relation connecting the principal values of tan"'* and 
cot"ia;: 

(15) tan-* X + cot"* « = f. 
Hence 

(16) lco1r>a; = ^, 

^ ' dx 1 + 3^' 

or 

(17) ci!cot-ia; = ^. 

It is well for the student to make a graph of this function, 
also, drawing in the principal value, as usual, in red. 

The following identity holds for positive values of x, when 
the principal values of the functions are used : 

(18) tan-i- = cot-i x, 0<x. 

X 

For negative values of x it reads : 

(180 tan-ii = cot-i x-tt, x<0. 

X 

Remarks. The other inverse trigonometric functions, sec-'o!, 
cacr^x, can be treated in a similar manner. They are, how- 
ever, without importance in practice. Their principal values 
cannot be defined by means of a single continuous curve. 
The graph necessarily consists of more than one piece ; it is 
most natural to take it as consisting of two pieces. 

Corresponding to the Addition Theorem for each of the 
trigonometric functions, there are functional relations for the 
inverse trigonometric functions. Thus, for tan-'a! : 

(19) tan-i u + tan-i v = tan-i '" + ^ • 
^ ' 1 — WW 

These relations, however, are not always true when the 
principal value of each of the functions is taken, and for this 



THE INVERSE TRIGONOMETRIC FUNCTIONS 215 

reason it is usually better not to employ them. If, however, 
ia a particular case, u and v are each numerically less than 
unity, the principal values can be used throughout in (19). 

3. Shop Work. The student will now add to his list of 
Special Formulas the four new formulas of this chapter. The 
list of formulas of dififerentiation is now complete. It reads 
as follows. 

Special Formulas of Differentiation 

1. dc = 0. 

2. d a" = na;"~' da;. 

3. d sin x = cos x dx. 

4. d cos x = — sin x dx. 

5. dtan X = sec' x dx. 

6. d cot x = — esc' X dx. 

7. 



8. 
9. 

10. 



11. 



12. 



13. 



dloga; 


X 




de" 


= e^da;. 




da" 


= a' log a 


da;. 


d sin~^ X 


dx 






Vl-a;2 




d cos~^ X 


dx 






VI- 


^ 


d tan"' SB 


da; 
1+ a;2 




d cot-la;: 


da; 





l + a;2 



It is important that the student gain facility in the use of 
the new results. 



216 CALCULUS 

Exa/mple 1. Differentiate the function 

u = cos~i - , ■ a > 0. 



Let y =s 



a 
Then u = cos"' y, 

du = d cos~i y 

— ^y 



Hence — - 



, dx 
dy = — 
a 

dx 
dy a dx 



VI — f r. 7«Y Va2 - X' 



and, finally, 



dC08-^ = - ''«' 



« Va2 — a!2 
In abbreviated form. 



«) 



4 , 

a cos""^ - = - 



Example 2. Differentiate the function 



Va2 — ! 









« = tan-i2«' + l. 
3 




Here, 


d 


20! + 1 










3 


idx 


3dx 




1 + 


r2x + v 

K 3 , 


f 10 + 4x4-43!!' 

/ 9 


^5+2x + 2!ifi 



du 
or 



dx 5 + 2x + 2a^ 



THE INVERSE TRIGONOMETRIC FUNCTIONS 217 

The student should notice that the method used in the text 
for deriving the fundamental formulas of differentiation is not 
to he repeated in the applications. It is these formulas them- 
selves that should be used. Thus, to solve Ex. 1 by writiag 

X 

cos u = - 
a 

and then differentiating would be logically irreproachable, but 
bad technique. 

EXERCISES 

Differentiate each of the following functions. 

1. M = smi — — iH-. --^11 

a 

2. M = tan~i — 

a 

3. M = cot~'-- 

a 

4. M = sin~i(w sin x). 

-1 1 — « 

5. M = cos 1 — - — • , , 

2 dx V3-|-2a; — a;2 

. ,2a! — 1 „ i.-,x+a 

6. M = sm-i=^^^^ — =• 7. ?t = coti--f — 

V2 * 

8. M = cot-1 = • 9. M = tan-'-- 

X 

10. u = tan ' 

1 — X- 



dx 


V^ 


— x- 


du 


1 




dx 


a^ + 


X"' 


du = 


''~a^ 


dx 

+ x^ 


du 




ncosa; 


dx 


vr 


— in? sitf X 


du 




1 



;=:tan-V a;-— ; 



11. u^tssr^ix^ — — -„ 
■ 3a;' 



■ _,x — a 
12. M = sm 1 







X 






du 


2 






dx 


l-|-a;2 






du 


3 






dx~ 


1 + a;' 




13 


. u = 


: C0S"1 — 
X 


X 

+ a 




du 


2 





^ dx Vl — a;* V2 



218 

15. « = C0S-1-. 

2 

16. t=sin-ii. 



17. «=cos-i-+Y. 
n 



18. u= X sin^i a;. 
1 



CALCULUS 




^ = -2siii2t 
dt 




- = 3cos3<. 
(if 




ds . . 
— = — nsiii?i(<- 


-y)- 


19. „-tan-ix 

X 




du 1 





20. u = - _ ^ . 

^^^^ <^ Vr;=^(siii-ix)2 

21. M = acos-i^^l^. 22. M = tan-i ^~" . 

a rc + a 

23. M = cot-i^±^. 24 «=.sm-i55^±^. 
03! — a bx + a 



25. M=V^^^a^-acos-i^. *f = X?lEZ, « > 0. 

a; dx X 

26. w = sm-i? + ^^^H^ ** = ^^EZ,a>0 

ax dx x^ 

du 2 



27. M = tan-i/'2tan-^ 

28. M = tan-i(3 tan 6). 



da; 5 — 3 cos x 
du 3 



de 5 — 4cos2i9 



4. Continnation. Numerical Computation. By means of 
the Tables the numerical value of any of the functions of this 
chapter can be determined when a specific numerical value has 
been chosen for the independent variable. It is, however, an 
important aid to ease and security in such computations to be 
able, in advance, to make sure of the early significant figures 
and the location of the decimal point. There are two impor- 
tant geometrical methods for achieving this end. One js the 
representation of the trigonometric functions by suitable lines 
connected with the unit circle ; the other consists in the graphs 
introduced above, in § 2. 



THE INVERSE TRIGONOMETRIC FUNCTIONS 219 

iFirBt of all, however, it should be pointed out that there are 
two distinct problems. One is to find all values of x which 
satisfy such equations as 

(a) sin a; = .2318 ; 

(6) cos a; = - .4322 ; 

(c) tan a; = - 1.4861. 

The other is to find the principal value of an inverse trigono- 
metric function ; for example, 

sin-1.2318; cos-i (- .4322) ; tan-i(- 1.4861) 

The methods of treating these problems are identical- 

First Geometric Method. Equations (a), (6), (c) can be 
solved graphically by the aid of the unit circle representation 
with an error corresponding to a degree or two, the results 
being expressed in radians if the problem comes from the 
Calculus. 

For example, consider equation (6). The student should 
provide himself with an accurately drawn circle of his own 
construction, executed on the accurate centimeter-millimeter 
paper commercially procurable ; the radius of the circle berug 
10 cm. and its center at a principal intersection of the rulings. 

To solve equation (6), he will lay a straight-edge on his 
plate, parallel to the secondary (or y-) axis and at a distance 
of 4 cm., 3^ mm. to the left of that axis. Marking the two 
points of intersection of the straight-edge with the circle by 
fine pencil lines easUy erased, he now measures one of the 
acute angles involved by means of his protractor and thus 
determines the two solutions of (6) lying between 0° and 360° 
correct to minutes or thereabouts. By aid of the Tables the 
values can at once be converted into radian measure. 

Arithmetic Solutions. From the figure before him the stu- 
dent now sees clearly a right triangle, one leg of which is 
known. The determination of the angle he needs is merely a 
problem in the solution of a right triangle by the tables, and 



220 CALCULUS 

he proceeds to carry this work through to the degree of accu- 
racy which the tables permit. 

Equations (a) and (c) are treated in a similar manner. The 
point of this method is that the student is trained to visualize 
a figure, and not to try to remember a table that looks like 

sin^ + + — — . 

For, such tables vanish ia a short time, and when the student 
needs his trigonometry ia later work, he is helpless. 

In terms of the inverse functions, this first problem consists 
in finding all the values of the multiple-valued function cos~i x 
for the value of the variable, x = — .4322. 

Second Oeometric Method. This method consists ia readiag 
off from the graph the two values which lie between and 2 jr, 
and then adding to these arbitrary positive or negative multi- 
ples of 2 jr. 

The graph suggests, moreover, how to determine these values 

arithmetically by the aid of a table of sines or cosines of angles 

of the first quadrant. It also suggests a further refinement of 

the graphical method, of which the student will do well to 

avail himself, — namely, this. Let him make an accurate 

graph of the function 

° ^ y— sin X 

on cm.-mm.-paper, taking 10 cm. as the unit and measuring 
the angle in radians, x ranging from to ir/2. This half-arch 
supplements the four graphs of the functions sin x, cos x, sin~i x, 
cos~ia; and serves as a 3-place table for determining their values 
(with a possible error of two or three units in the third place). 

To sum up, then, there are two geometric methods ; 1) the 
unit-circle method ; 2) the graphs of the functions, the latter 
beiag supplemented by the 10-cm. graph just described. Either 
of the geometric methods suggests how to use the tables correctly 
and affords an altogether satisfactory check on the tables. 

When the accurately drawn graphs are not at hand, free- 
hand drawings indicate clearly how to use the tables with 
security and accuracy. 



THE INVERSE TRIGONOMETRIC FUNCTIONS 221 

EXERCISES 

1. Determine both in degrees and radians all values of x 
which satisfy the above equations (a), (6), (c), using each time 
all of the geometric methods set forth, and also the tables. 

2. Mnd the value of each of the following functions. It is 
understood that the principal value is meant. Use first the 
method of the graphs. Then determine from the tables. 
Check by unit-circle and protractor. 

i) sin-i(-.1643); ii) cos-i (.6417) ; 

iii) tan-i(- 2.8162). 

3. By means of a free-hand drawing of the graph estimate 
the value of each of the following functions. Remember that 
a curve recedes from its tangent very slowly near a point of 
inflection. 

a) sin-1.113; 6) tan-i(- .214) ; c) cos-i.l72; 

d) tan-i (- 7.4) ; e) cot"! ( - .162) ; /) cos-i (- .998) ; 

g) sin-i(-.21); h) sin-i.89; i) tan-i6.2; 

j) cot-17.3; fc) cos-i(-.138); Z) sin-i (- .138). 

In what cases is your error large ; in what, small ? 

5. Applications. The inverse trigonometric functions afford 
a convenient means of solving the following problem in Optics. 

A ray of light is refracted in a 
prism. Show that its deviation from 
its original direction is least when 
the incident ray and the refracted 
ray make equal angles with the 
faces of the prism. 

The study of this problem has a pj^ ^3 

vivid interest for the student who 

has seen the laboratory experiment of admitting a ray of 
sunlight into a darkened room, allowing it to pass through 




222 CALCULUS 

a prism, thus being refracted, and throwing it finally, dis- 
persed, on a screen. 

Let AF be the incident ray; PQ, its path through the 
prism ; and Q£ the ray -which emerges. Then the deflection 
of PQ is obviously 6— ^ and the further deflection of Q,B is 
6' — <l>' ; so that the total deflection, u, is : 

(1) u = e-4, + e'-4>' = e+6'-(ct> + <i>'). 

On the other hand, the sum of the angles of the triangle 
PDQ is / \ / \ 

'=(i-*Hi-*')+- 

Hence 

(2) ^ + 0'=a. 

We can, therefore, write (1) in the form : 

(3) u = e+e'-a. 

This is the quantity it is desired to make a minimum. and 
6' are, however, connected by a relation which can be obtained as 
follows. We have by the law of refraction (cf. Chap. V, § 7) : 



/A\ sin 5 

(') sin^-'^' 




sin 6' 

-: ;= n. 

sin<^' 


Letv=l/M. Then 






(6) sin <^ = V sin fl 


or 


<l> = sin~i (v sin 8). 


Similarly, 






(6) sin <^' = i/ sine' 


or 


^' = sin-i(i/sine') 



Substituting these values of ^ and </>' in equation (2) we have 
the desired relation : 

(7) sin-i (v sin ff) + sin-i (v sin 0') = a. 

Our problem now is completely formulated ; it is : To make 
the function u given by (3) a minimum, when 8 and 0' are con- 
nected by (7) : 

u = + 0'-a, 

sin"i (v sin 0) + sin~i (v sin 0') =a. 



(8) 



THE INVERSE TRIGONOMETRIC FUNCTIONS 223 
Take 6 as the independent variable. Then 

^' M-^ + W 

and the condition 

— - = gives -^ = — 1. 

de ^ d0 

Next, take the differential of each side of the second equar 
tion (8) : 

d(vsing) d(vsiD.6') ^ 

Vl — y2 sin2 d Vl — 1/2 sia2 6' 
or 

V cos 6dd , V cos ff d6' _ « 



Hence 



VI — 1/2 giji2 Q VI _ y2 sui2 01 

(■\n\ cos g . cos^^ /'^^''\_o 

Vl - v2 sin2 6 Vl-i/2sin2 0'Uey 



But dO'/dO = — 1. Consequently 

/-. -| ^ cos 5 cos 6' 



VI — v2 sin2 a Vl - v^ sin2 e' 

One solution of this equation is 6 = 6', — the solution de- 
manded by the theorem. But conceivably there might be 
other solutions, and then it -would not be clear which one of 
them makes u a miaimum. We can readUy show, however, 
that equation (11) has no further solutions. Square each side : 

cos^g ^ cos^y 
l-v2siii2e l-v-'Sia^e'' 

Clear of fractions and express each cosine in terms of the sine : 

(1 - v2 sin2 0')(1 - sin' S)=(l- v"- sitf e)(l - sia^ 6'). 

Multiply out and suppress equal terms on the two sides : 

_ sin2 e - v"' sin2 & = - sin' ff — v^ sin' B, 

if - 1) sin2 e = (v2 - 1) sin2 6'. 



224 CALCULUS 

Hence 

siii2 e = siii2 fl', sin 6 = sin 6', 

and consequently the only angles of the first quadrant which 
can satisfy (11) are equal angles, 6 = 6'. 
From (5) and (6) it follows that <f> = <^'. Hence, from (2) 

(^ = ^, and so $= sin-i( n sin^j, 
M = 2sin"i( Msin- )— a. 

\ V 

That M is a minimum, is clearly indicated by the labora- 
tory experiment. It can be proven analytically as follows. 
From (9) 

dW- dfl2 ' 

Differentiate (10) as it stands ; then, after the differentia- 
tion, set dd'/dd = — 1 and & = 6'. It is seen at once tha*- 

^'<0, hence ^ > 0, 

and u has a minimum. 

EXERCISE 

The bottom of a mural painting 4 ft. high is 12 ft. above 
the eye of the observer. How far back from the wall should 
he stand, in order that the angle subtended by the painting be 
as large as possible ? 

Suggestion. Take the distance, x, of the observer from the 
wall as the independent variable, and express the angle of 
elevation of the bottom and the top of the painting in terms 
of X. 



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